text:
EuroLex/F/General

 * Original language: French
 * Original form and meaning: général - 1. general, commander, military leader

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: shortening of "captain general", from MF capitaine général. Orignally derived from L generalis (adj.) "general", from L genus (nom.) "gender, kind, category", came into army use under French influence in the 16th century.

Source: http://www.etymonline.com, http://www.koeblergerhard.de/derwbhin.html

Information on Other Languages
Hungarian: tábornok

title: EuroLex/F/General


text:
EuroLex/F/Grave accent

 * Original language: French
 * Original form and meaning: accent grave - 1. a diacritic mark placed above a vowel to indicate pronunciation

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
The word grave is derived from the Latin "gravis" (heavy), itself a translation of the Greek "barys" (βαρύς). In English the word is normally pronounced "grahv", IPA /ɡɹɑːv/, not like "grave" meaning serious or a tomb. In French it is pronounced similarly: accent grave /aksɑ̃ ɡʁav/.

Source: http://www.answers.com

Information on Other Languages
-

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

title: EuroLex/F/Grave accent


text:
EuroLex/F/Haute couture

 * Original language: French
 * Original form and meaning: from haut, haute + couture - 1. fashion tailoring, couture

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Haute couture


text:
EuroLex/F/Haute cuisine

 * Original language: French
 * Original form and meaning: from haut, haute + cuisine  - 1. grand cuisine

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: 1926, from F, literally "high(-class) cooking"

Source: http://www.etymonline.com

Information on Other Languages
-

title: EuroLex/F/Haute cuisine


text:
EuroLex/F/Jacket

 * Original language: French
 * Original form and meaning: jaquette 'book wrapper, coat'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Jacket


text:
EuroLex/F/Jalousie

 * Original language: French
 * Original form and meaning: jalousie - 1. jalousie, blind

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from F, literally "jealousy", from the notion of looking through blinds without being seen

Source: http://www.etymonline.com

Information on Other Languages
-

title: EuroLex/F/Jalousie


text:
EuroLex/F/Jardinière

2. gardener
 * Original language: French
 * Original form and meaning: 1. flower box

3. window box

4. vegetable platter

5. a dish garnished with vegetables, which are served in individual groups arranged around the main dish

6. jardiniere (a large decorative stand or pot for plants or flowers)

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
French, from feminine of jardinier - gardener, from Old French, jardin - garden

source: http://dictionary.reference.com

Information on Other Languages
-

title: EuroLex/F/Jardinière


text:
EuroLex/F/Jargon

 * Original language: French
 * Original form and meaning: jargon - 1. terminology; 2. cant, gibberish, mumbo-jumbo; 3. pidgin

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from OF jargon "a chattering" (of birds), ultimately of echoic origin (cf. L. garrire "to chatter", English gargle). Often applied to something the speaker does not understand, hence meaning "mode of speech full of unfamiliar terms" (1651).

Source: http://www.etymonline.com

Information on Other Languages
meaning 1

Czech: slang, hantýrka

Estonian: erikeel

Finnish: erikoiskieli

title: EuroLex/F/Jargon


text:
EuroLex/F/Jealous

 * Original language:French
 * Original form and meaning:jaloux 'envious'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Jealous


text:
EuroLex/F/Jelly

 * Original language: French
 * Original form and meaning: gelée - 1. frost, freeze; 2. jelly, aspic

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from OF gelée "a frost, jelly", fem. pp. of geler "congeal", from L gelare "to freeze", from gelu "frost".

Jellyfish as the popular name of the medusa and similar sea-creatures is from 1841. Jellybean first attested 1908. Jellyroll "cylindrical cake containing jelly or jam" is from 1895; as slang for "vagina, sexual intercourse" it dates from 1914.

Source: http://www.etymonline.com

Information on Other Languages
meaning 2:

Estonian: tarretis

Finnish: hyytelö

title: EuroLex/F/Jelly


text:
EuroLex/F/Journal

 * Original language: French
 * Original form and meaning: journal - 1. leaf, page, sheet; 2. magazine, paper; 3. newspaper; 4. diary, (nautical) log

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from Anglo-F jurnal "a day", from OF journal, originally "daily" (adj.), from LL diurnalis "daily" c.1355, "book of church services". Sense of "daily record of transactions" first recorded 1565; that of "personal diary" is 1610, from a sense found in French.

Source: http://www.etymonline.com

Information on Other Languages
meaning 2

Czech: časopis

Danish: tidsskrift

Dutch: tijdschrift

Estonian: ajakiri

Finnish: aikakauslehti

Hungarian: folyóirat

Italian: rivista

Norwegian: tidsskrift

Polish: czasopismo

Portuguese (Portugal): revista

Slovak: časopis

Slovenian: revija

Spanish: revista

Swedish: tidskrift, tidning

meaning 4

Czech:	deník

Danish: dagbog

Dutch: dagboek

Estonian: päevik

Finnish: päiväkirja

Hungarian: napló

Italian: diario

Latvian: dienasgrāmata

Norwegian: dagbok, logg(bok)

Polish: dziennik

Slovak: denník

Slovenian: dnevnik

Spanish: diario

title: EuroLex/F/Journal


text:
EuroLex/F/Journalism

 * Original language: French
 * Original form and meaning: journalisme - 1. journalism

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from Anglo-F jurnal "a day", from OF journal, originally "daily" (adj.), from LL diurnalis "daily". Journalism is 1833 in Eng, likewise from F (where it is attested from 1781).

Source: http://www.etymonline.com

Information on Other Languages
Hungarian: újságírás

Polish: dziennikarstwo

Slovenian: novinarstvo

Spanish: periodismo

title: EuroLex/F/Journalism


text:
EuroLex/F/Journalist

 * Original language: French
 * Original form and meaning: journaliste - 1. journalist

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from Anglo-F jurnal "a day", from OF journal, originally "daily" (adj.), from LL diurnalis "daily"

Source:http://www.etymonline.com

Information on Other Languages
Czech: male: novinář, female: novinářka Estonian: ajakirjanik

Hungarian: újságíró

Polish: dziennikarz Slovak: male: novinár, female: novinárka

Slovenian: novinar

Spanish: periodista

title: EuroLex/F/Journalist


text:
EuroLex/F/Lemon

 * Original language: French
 * Original form and meaning: limon - 1. lemon (Citrus limon); 2. [techn.] drawbar; 3. mud, [geol.] silt

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from OF limon "citrus fruit", from OProv, from Ar laimun or Pers limu(n), [...] cognate with Skt. nimbu "the lime".

Slang meaning "a Quaalude" is 1960s, from Lemmon, name of a pharmaceutical company that once manufactured the drug.

AmE slang "worthless thing", 1909, from lemon, perhaps via criminal slang sense of "a person who is a loser, a simpleton", which is perhaps from the notion of someone a sharper can "suck the juice out of".

To hand someone a lemon was British slang (1906) for "to pass off a sub-standard article as a good one". Or it simply may be a metaphor for something which "leaves a bad taste in one's mouth".

Source: http://www.etymonline.com

Information on Other Languages
meaning 1:

Czech: citrón, citrónový

Danish: citron

Dutch: citroen

Estonian: sidrun

Finnish: sitruuna

German: Zitrone

Hungarian: citrom

Latvian: citrons

Lithuanian: citrina

Norwegian: sitron

Polish: cytryna

Slovak: citrón, citrónový

Swedish: citron

meaning "lime":

Lithuanian: rūgščiavaisio citrinmedžio vaisius

Slovenian: citronka

title: EuroLex/F/Lemon


text:
EuroLex/F/Lemonade

 * Original language: French
 * Original form and meaning: limonade - 1. lemonade

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from OF limon "citrus fruit", from OProv, from Ar laimun or Pers limu(n) [...], cognate with Skt nimbu "the lime". Lemonade is first recorded 1663, from F limonade; earlier E spelling was lemonado (c.1640) with false Sp ending.

Source: http://www.etymonline.com

Information on Other Languages
Czech: citrónovka

Portuguese (Portugal): gasosa

Slovak: citronáda

title: EuroLex/F/Lemonade


text:
EuroLex/F/Let-pass

 * Original language:French
 * Original form and meaning:laissez-passer 'free pass'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Let-pass


text:
EuroLex/F/Lieutenant

 * Original language: French
 * Original form and meaning: lieutenant - 1. lieutenant

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from OF lieu tenant "substitute", literally "placeholder", from lieu "place" + tenant, prp. of tenir "to hold". The notion is of a "substitute" for higher authority. Specific military sense of "officer next in rank to a captain" is from 1578.

Pronunciation with lef- is common in Britain, and spellings to reflect it date back to 14c., but the origin of it is mysterious.

Source: http://www.etymonline.com

Information on Other Languages
Czech: nadporučík

Hungarian: főhadnagy, (Amerikában:) hadnagy

Italian: tenente

Polish:	porucznik

Portuguese: tenente

Slovak: nadporučík

Slovenian: poročnik

Spanish: teniente

title: EuroLex/F/Lieutenant


text:
EuroLex/F/Limousine

 * Original language: French
 * Original form and meaning: limousine - 1. limousine (i.e. luxury car)

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from F limousine, from Limousin, region in central France, originally an adjective referring to its chief city, Limoges.

Modern automobile meaning evolved from perceived similarity of the car's profile to a type of hood worn by the inhabitants of that province. Since 1930s, synonymous in AmE with "luxury car"; applied from 1972 to vehicles that take people to and from large airports. Abbreviation limo is 1960s, AmE Limousine liberal first attested 1969.

Source: http://www.etymonline.com

Information on Other Languages
-

title: EuroLex/F/Limousine


text:
EuroLex/F/Livery

 * Original language: French
 * Original form and meaning: livrée 'service clothing'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Livery


text:
EuroLex/F/Macabre

 * Original language: French
 * Original form and meaning: macabre - 1. macabre

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from OF (danse) Macabré "(dance) of Death" (1376), probably a translation of ML (Chorea) Machabæorum, literally "dance of the Maccabees" (leaders of the Jewish revolt against Syro-Hellenes). The association with the dance of death seems to be via vivid descriptions of the martyrdom of the Maccabees in the Apocryphal books. The abstracted sense of "gruesome" is first attested 1842 in F, 1889 in Eng.

Source: http://www.etymonline.com

Information on Other Languages
Czech: strašidelný Dutch: griezelig

Estonian: õudne

Hungarian: hátborzongató Latvian: drausmīgs, šaušalīgs

Slovak: strašidelný

Slovenian: grozoten, obešenjaški

title: EuroLex/F/Macabre


text:
EuroLex/F/Maisonette

 * Original language: French
 * Original form and meaning: maisonnette - 1. a small house, a cottage

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
maisonnette: diminutive of "maison"

Information on Other Languages
The French term for a flat within a larger house or block, especially one on two floors, usually with its own separate entrance is "appartement en duplex".

The Spanish term for a flat within a larger house or block, especially one on two floors, usually with its own separate entrance is "dúplex".

title: EuroLex/F/Maisonette


text:
EuroLex/F/Marriage

 * Original language:French
 * Original form and meaning:mariage 'wedding'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Marriage


text:
EuroLex/F/Massage

 * Original language: French
 * Original form and meaning: massage - 1. massage

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from F massage "friction of kneading", from masser "to massage", possibly from Ar massa "to touch, feel, handle"; if so, probably picked up in Egypt during the Napoleonic campaign there. Other possibility is that F got it in colonial India from Port amassar "knead", a verb from L massa "mass, dough".

The verb is attested from 1887.

Masseur, fem. masseuse, both dating to 1876, are from Fr. masser. Massage parlor first attested 1913, from the start a euphemism for "house of prostitution".

Source: http://www.etymonline.com/

Information on Other Languages
Finnish: noun: hieronta; verb: hieroa

title: EuroLex/F/Massage


text:
EuroLex/F/Menage

 * Original language: French
 * Original form and meaning:ménage 'household, housekeeping, cleanup (of house)'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
- OED: The Oxford English Dictionary: Second Edition. Clarendon Press, Oxford 1989

Information on Other Languages
-

title: EuroLex/F/Menage


text:
EuroLex/F/Meringue

 * Original language: French
 * Original form and meaning: meringue - 1. meringue (in the sense of: a topping for pastry or pies made of a mixture of egg whites and sugar beaten until stiff and often baked until brown or a small pastry shell or cake made of stiffly beaten, baked egg whites and sugar, often containing fruit or nutmeats)

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: unknown origin

Perhabs to be identified with dialect (Walloon) maringue "shepherd's loaf", marinade food for an outdoor repast (< L merenda "light afternoon meal", probably fem. ger. of merere "to merit", such a meal being part of a labourer's wages). Certain evidence is lacking. association with the town of Meiringen (Bern canton, Switzerland) is solely by folk etymology.

Sorce: http://www.etymonline.com, http://dictionary.reference.com

Not to be mixed up with merengue (sometimes also calles méringue; is of AmerSp origin: 1935–40) : noun - 1. a ballroom dance of Dominican and Haitian origin, characterized by a stiff-legged, limping step

2. the music for this dance

verb (used without object) - to dance the merengue

Source: http://dictionary.reference.com

Information on Other Languages
all meaning 1:

Czech:                sněhová pusinka

Dutch:                schuimgebakje

Estonian:             meering, besee

German (more common): Baiser (also of French origin baiser "kiss", this from Latin basiare "to kiss"): meaning 1, and also a male first name from Frisia (Northern Germany)

Hungarian:            habcsók

Latvian:              bezē

Lithuanian also:      bezė

Polish:               beza

Slovak:               snehová (cukrová) pusinka

Slovenian:            pecivo iz beljakove pene

title: EuroLex/F/Meringue


text:
EuroLex/F/Migraine

 * Original language: French
 * Original form and meaning: migraine - 1. migraine, headache

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: ME megrim, from OF migraigne, from vulgar pronunciation of LL hemicrania "pain in one side of the head, headache", from Gk hemikrania, from hemi- "half" + kranion "skull". The ME form was re-spelled 1777 on F model.

Source: http://www.etymonline.com; http://www.allwords.com

Information on Other Languages
Portuguese: enxaqueca

title: EuroLex/F/Migraine


text:
EuroLex/F/Musket

 * Original language: French
 * Original form and meaning: mousquet - 1. musket

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from MF mousquette, a kind of sparrow-hawk, diminutive of mosca "a fly", from L musca. The hawk so called either for its size or because it looks speckled when in flight. Early firearms were often given names of beasts (cf. dragoon), and the equivalent word was used in It to mean "an arrow for a crossbow". The MF word was borrowed earlier (c.1425) in its literal sense of "sparrow-hawk".

Musketeer "soldier armed with a musket" is 1590, from F mousquetaire, from mousquette.

Source: http://www.etymonline.com

Information on Other Languages
-

title: EuroLex/F/Musket


text:
EuroLex/F/Niche

 * Original language: French
 * Original form and meaning: niche - 1. gap, cavity, recess; 2. niche (social, personal, natural); 3. market gap; 4.dog kennel

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology:

Source: http://www.etymonline.com

Information on Other Languages
meaning 1:

Finnish: syvennys

Estonian: orv

Hungarian:  fülke

meaning 4:

Czech: místečko, pevné místo

Danish: rette hylde

Dutch: plaats

Estonian: (oma) koht

Finnish: elämäntehtävä

Hungarian: hely

Italian: posto

Latvian: īsta

Lithuanian: tinkama vieta

Polish: odpowiednie miejsce

Portuguese: lugar

Slovak: vhodné miesto

Slovenian: mesto

Spanish: hueco

title: EuroLex/F/Niche


text:
EuroLex/F/Parasol

 * Original language: French
 * Original form and meaning: parasol 'sunshade'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Parasol


text:
EuroLex/F/Patrol

Cf. also Patrouille (EuroLex).

 * Original language: French
 * Original form and meaning: patrouille 'patrol, reconaissance patrol'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Patrol


text:
EuroLex/F/Patrouille

Cf. Patrol (EuroLex).

 * Original language: French
 * Original form and meaning: patrouille 'patrol, reconaissance patrol'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Patrouille


text:
EuroLex/F/Perfume

 * Original language: French
 * Original form and meaning: parfum - 1. flavour; 2. aroma, scent; 3. perfume; 4.taste additive

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from MF parfum, from parfumer "to scent", from Prov. perfumar, from L per- "through" + fumare "to smoke". Earliest use in English was in reference to fumes from something burning. Meaning "fluid containing agreeable essences of flowers, etc., is attested from 1542.

Source: http://www.etymonline.com

Information on Other Languages
meaning 3: Czech: voňavka

Estonian: lõhnaaine, lõhnaõli

Latvian: smaržas

Lithuanian: kvepalai

Slovak: voňavka

title: EuroLex/F/Perfume


text:
EuroLex/F/Petit

 * Original language: French
 * Original form and meaning: petit 'small, little, narrow, young, humble'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Petit


text:
EuroLex/F/Petite

 * Original language: French
 * Original form and meaning: petit 'small, little, narrow, young, humble'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Petite


text:
EuroLex/F/Praline

 * Original language: French
 * Original form and meaning: praliné 'little sweets made of chocolate'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Praline


text:
EuroLex/F/Press

 * Original language:French
 * Original form and meaning:presser 'to push, to urge'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Press


text:
EuroLex/F/Pressing

 * Original language:French
 * Original form and meaning:presser 'to push, to urge'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Pressing


text:
EuroLex/F/Ragout

 * Original language: French
 * Original form and meaning: ragoût 'stew, hot pot'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Ragout


text:
EuroLex/F/Sadism

 * Original language: French
 * Original form and meaning: sadisme - 1. 'The deriving of sexual gratification or the tendency to derive sexual gratification from inflicting pain or emotional abuse on others.'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
After Count Donatien A.F. de Sade (1740-1815). Not a marquis, though usually now called one, he was notorious for the cruel sexual practices he described in his novels. (source: http://www.etymonline.com)

Information on Other Languages
Russian: садизм

Greek: σαδισμός

(source: http://www.answers.com)

title: EuroLex/F/Sadism


text:
EuroLex/F/Terrace

 * Original language: French
 * Original form and meaning: terrasse 'patio, loft'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

 * Annotations: -

 * Information on Other Languages: -

title: EuroLex/F/Terrace


text:
EuroLex/F/Toilet

 * Original language: French
 * Original form and meaning: toilette 'lavatory, bathroom, to dress up'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: original English meaning: "cover or bag for clothes", from MF toilette "a cloth, bag for clothes" diminutive of toile "cloth, net". Sense evolution is to "act or process of dressing" (1681); then "a dressing room" (1819), especially one with a lavatory attached; then "lavatory or porcelain plumbing fixture" (1895), an AmE euphemistic use. Toilet paper is attested from 1884. Toilet training is recorded from 1940.

Source: http://www.etymonline.com

Information on Other Languages
Estonian: WC

Finnish: WC

Portuguese: privada, banheiro

Slovenian: stranišče

Spanish: lavabo

title: EuroLex/F/Toilet


text:
EuroLex/F/Toilet water

 * Original language: French
 * Original form and meaning: eau de toilette - 1. toilet water (a scented liquid with a high alcohol content used in bathing or applied as a skin freshener, similar to cologne)

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology of Toilet (EuroLex).

Information on Other Languages
loan-translations:

German: Toilettenwasser

Lithuanian: tualetinis vanduo (?)

Polish: woda toaletowa

Slovenian: toaletna voda

title: EuroLex/F/Toilet water


text:
EuroLex/F/Toilette

Cf. also Toilet (EuroLex).

 * Original language: French
 * Original form and meaning: toilette 'lavatory, bathroom, to dress up'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Toilette


text:
EuroLex/F/Trace

 * Original language: French
 * Original form and meaning: trace 'tincture, spoor, track, hint'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Trace


text:
EuroLex/F/Vis-à-vis

 * Original language: French
 * Original form and meaning: vis-à-vis 'face to face, counterpart, opposite number, love seat'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Vis-à-vis


text:
EuroLex/F/Visage

 * Original language:French
 * Original form and meaning:visage 'face, expression, mime'

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/Visage


text:
EuroLex/F/Wardrobe

 * Original language: French
 * Original form and meaning: derived from the French word Garderobe (EuroLex)

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
Etymology: from ONF warderobe, variant of OF garderobe "place where garments are kept", from warder "to keep, guard" + robe "garment". Meaning "a person's stock of clothes for wearing" is recorded from c.1400. Sense of "movable closed cupboard for wearing apparel" is recorded from 1794. Meaning "room in which theatrical costumes are kept" is attested from 1711.

Source: http://www.etymonline.com, http://dictionary.reference.com

Information on Other Languages
-

--Assistant1 15:10, 7 February 2007 (CET)

title: EuroLex/F/Wardrobe


text:
EuroLex/F/À la jardinière

 * Original language: French
 * Original form and meaning: à la jardinière - 1. a dish garnished with vegetables, which are served in individual groups arranged around the main dish

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Information on Other Languages
-

title: EuroLex/F/À la jardinière


text:
EuroLex/F/À la maison

 * Original language: French
 * Original form and meaning: à la maison - 1. at home,

2. in combination with verbs: rentrer à la maison (send home); arriver à la maison (arrive at home),

3. culinary: a designation like "pâté maison" refers to a specialty of the house or to the fact that the dish was made by the house chef

(Note: If the status is not specifically indicated then the word is stylistically neutral and generally used; if earlier meaning and status equals current use the former may be expressed by writing "dito". Cf. also the project guidelines.)

Annotations
-

Information on Other Languages
-

title: EuroLex/F/À la maison


text:
User:Eml5526.s11.team7.flater/EGM6611.HW3

=Homework 3=

Description
Derive the conservation of mass equation.

Solution

 * Adapted from Malvern, Section 5.2.

Consider an arbitrary volume V bounded by the surface S. If a continuous medium of density $$\rho$$ fills the volume at time t, the total mass in V is

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$$ \displaystyle M=\int_V \rho dV $$. (1.1)
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Since density is a function of position and time, the rate of increase of the total mass in the volume is the partial derivative

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$$ \displaystyle \frac{\partial M}{\partial t}=\int_V \frac{\partial \rho}{\partial t}dV $$.     (1.2)
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From the conservation of mass we assume that the rate of increase in V is equal to the rate of inflow of mass through the surface S. The mass flux inflow per unit area (dS) is defined as

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$$\displaystyle \text{Flux} = -\int_S \rho \mathbf{v}\cdot \mathbf{\hat{n}}dS $$.     (1.3)
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 * }

Recall that the divergence theorem states that the integral of the outer normal component of a vector over a closed surface is equal to the integral of the divergence (designated $$\displaystyle \nabla $$ or "nabla") of the vector over the volume bounded by the closed surface.

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$$\displaystyle \int_S \mathbf{v}\cdot \mathbf{\hat{n}}dS = \int_V \mathbf{\nabla} \cdot \mathbf{v}dV $$.     (1.4)
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Therefore equations 1.3 and 1.4 can be combined and equated to 1.2.

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$$\displaystyle \int_V \left[ \frac{\partial \rho}{\partial t} + \mathbf{\nabla} \cdot (\rho \mathbf{v}) \right]dV=0 $$.     (1.5)
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For any arbitrary volume V the integral becomes

 * $$\displaystyle\frac{\partial \rho}{\partial t} (x,t) + \nabla (\rho \mathbf{v}) =0$$

 * or

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$$\displaystyle \frac{\partial \rho}{\partial t} + \rho \nabla \mathbf{v} =0 $$     (1.6)
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 *
 * }

Thus we are left with the Eulerian description of the mass conservation.

Description
Derive the balance of linear momentum equations.

Solution

 * Adapted from Malvern, Section 5.3.

Let us consider the same control volume from problem 1, but in this case the mass is acted upon by external surface and body forces - t being applied per unit area ($$\displaystyle dS$$) and b applied per unit mass ($$\displaystyle \rho dV$$). The momentum principle as defined above can then be written as

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$$\displaystyle \frac{d}{dt}\int_V \rho \mathbf{v}dV = \int_S \mathbf{t}dS + \int_V \rho \mathbf{b}dV $$     (2.1)
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 *
 * }

 * where the left-hand side of the equation represents the rate of change of the total momentum. In rectangular coordinates, equation 2.1 takes the form

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$$\displaystyle \frac{d}{dt}\int_V \rho \ v_i \ dV = \int_S t_i \ dS + \int_V \rho \ b_i \ dV $$ (2.2)
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 *
 * }

First, consider the definition of a traction vector in indicial notation: $$\displaystyle t_i = T_{ji}n_j$$. Second, recall the divergence theorem from equation 1.4. Now equation 2.2 can be expressed as

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$$\displaystyle \int_V \rho \frac{dv_i}{dt} dV = \int_V \left( \frac{\partial T_{ji}}{\partial x_j} + \rho b_i \right) dV \qquad \Rightarrow \qquad \int_V \left( \frac{\partial T_{ji}}{\partial x_j} + \rho b_i - \rho \frac{dv_i}{dt}\right) dV =0$$. (2.3)
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As in problem 1, for any arbitrary volume V the integral becomes

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$$\displaystyle \frac{\partial T_{ji}}{\partial x_j} + \rho b_i = \rho \frac{dv_i}{dt} $$     (2.4)
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 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *
 * }

 * or

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$$\displaystyle \mathbf{\nabla \cdot T} + \rho \mathbf{b} = \rho \frac{d\mathbf{v}}{dt} $$     (2.5)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *
 * }

Description
Derive the balance of moment of momentum equations.

Solution

 * Adapted from Malvern, Section 5.3.

Similar to the balance of linear momentum, the moment of momentum principle states that the time rate of change of the total moment of momentum is equal to the vector sum of the moments of the external forces acting on the system. In other words

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$$\displaystyle \frac{d}{dt}\int_V (\mathbf{r} \times \rho \mathbf{v}) dV = \int_S (\mathbf{r} \times \mathbf{t}) dS + \int_V (\mathbf{r} \times \rho \mathbf{b}) dV $$. (3.1)
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 *
 * }

The cross product of two vectors in a right-handed coordinate system is defined as $$\mathbf{c} = \mathbf{a} \times \mathbf{b} \quad \rightarrow \quad c_p = e_{pqr}a_q b_r = e_{qrp} a_q b_r $$. So 3.1 can be rewritten in indicial notation as

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$$\displaystyle \frac{d}{dt}\int_V (e_{rmn} x_m \rho v_n) dV = \int_S (e_{rmn} x_m t_n) dS + \int_V (e_{rmn} x_m \rho b_n) dV $$. (3.2)
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 *
 * }

Following the same procedure in problem 2 for equations 2.2 and 2.3, the surface integral can be transformed to a volume integral.

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$$\displaystyle \int_V e_{rmn} \frac{d}{dt} (x_m v_n) \rho \ dV = \int_V e_{rmn} \left[ \frac{\partial (x_m T_{jn})}{\partial x_j} + x_m \rho b_n \right] dV $$ (3.3)
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 *
 * }

Since $$\displaystyle \frac{dx_m}{dt}=v_m$$, equation 3.3 becomes

 * {| style="width:100%" border="0"

$$\displaystyle \int_V e_{rmn} \left( v_m v_n + x_m \frac{dv_n}{dt} \right) \rho \ dV = \int_V e_{rmn} \left[ x_m \left( \frac{\partial T_{jn}}{\partial x_j} + \rho b_n \right) + \delta_{mj}T_{jn}\right] dV $$. (3.4)
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 *
 * }

Note the following:
 * Since $$\displaystyle v_m v_n$$ is symmetric and $$\displaystyle e_{rmn}$$ is antisymmstric, $$\displaystyle e_{rmn} v_m v_n=0$$.
 * From the conservation of linear momentum (equation 2.4 above) the last term of the left-hand side cancels with the first term of the right-hand side.
 * $$\displaystyle \delta_{mj} T_{jn}=T_{mn}$$.

Therefore the only remaining terms of equation 3.4 are

 * {| style="width:100%" border="0"

$$\displaystyle \int_V e_{rmn} T_{mn} dV=0 $$.     (3.5)
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 *
 * }

For any arbitrary volume V the integral becomes

 * {| style="width:100%" border="0"

$$\displaystyle e_{rmn} T_{mn}=0 $$     (3.6)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *

 * }

 * proving that the stress matrix, $$\displaystyle T_{mn}$$, is symmetric.

 * {| style="width:100%" border="0"

$$\displaystyle T_{mn}=T_{nm} $$     (3.7)
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 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Description
Derive the conservation of energy equation.

Solution

 * Adapted from Malvern, Section 5.4.

The conservation of energy equations in continuum mechanics is derived from the first law of thermodynamics. In other words the total energy of the system ,E, is equal to the rate of heat, Q, and mechanical work, W, input to the system. The total energy is considered as sum of two parts, kinetic energy, K, and internal energy, U.

 * {| style="width:100%" border="0"

$$\displaystyle K + U = W + Q $$ (4.1)
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 *

 * }

 * where,

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$$\displaystyle K = \frac{d}{dt} \int_V \left( \frac{1}{2} \rho v_k v_k \right) dV $$, (4.2)
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 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle U = \frac{d}{dt} \int_V \rho u \ dV $$, (4.3)
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 *

 * }

 * {| style="width:100%" border="0"

$$\displaystyle W = \int_S \mathbf{t} \cdot \mathbf{v} \ dS + \int_V \rho \mathbf{f} \cdot \mathbf{v} \ dV $$, (4.4)
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 *

 * }

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$$\displaystyle Q = - \int_S \mathbf{q} \cdot \mathbf{\hat{n}} \ dS + \int_V \rho h \ dV $$. (4.5)
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 *

 * }

The heat input rate Q consists of conduction through the surface S, and a distributed internal heat source of strength h per unit mass. q is the heat flux vector. The power input (i.e. rate of work) is the rate at which the external surface tractions t per unit area and body forces b per unit mass are doing work on the system. Considering equation 4.4 in indicial notation and recalling the stress vector/tensor relationship (see equation 2.3), the surface integral is transformed to a volume integral. Equation 4.4 becomes

 * {| style="width:100%" border="0"

$$\displaystyle W = \int_V \left[ v_i \left( \frac{\partial T_{ji}}{\partial x_j} + \rho b_i \right ) + T_{ji} \frac{\partial v_i}{\partial x_j}\right ]dV $$     (4.6)
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 * style="width:10%; |
 *

 * }

 * and can be further reduced recognizing that the term inside the parentheses is equal to the balance of linear momentum equation (2.4).

 * {| style="width:100%" border="0"

$$\displaystyle W = \int_V \left[ v_i \left( \rho \frac{dv_i}{dt} \right ) + T_{ji} \ v_{i,j}\right ]dV \qquad = \qquad \frac{d}{dt} \int_V  \left( \frac{1}{2} \rho v_i v_i \right)dV + \int_V T_{ij}D_{ij}\ dV $$ (4.7)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

 * where $$\displaystyle D_{ij}$$ is the rate of deformation tensor.

Now equations 4.2, 4.3, 4.5, and 4.7 can be substituted into 4.1.

 * {| style="width:100%" border="0"

$$\displaystyle \frac{d}{dt} \int_V \left( \frac{1}{2} \rho v_k v_k \right) dV + \frac{d}{dt} \int_V \rho u \ dV = \frac{d}{dt} \int_V \left( \frac{1}{2} \rho v_i v_i \right)dV + \int_V T_{ij}D_{ij}\ dV - \int_S \mathbf{q} \cdot \mathbf{\hat{n}} \ dS + \int_V \rho h \ dV $$ (4.8)
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 *

 * }

This reduces to

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$$\displaystyle \frac{d}{dt} \int_V \rho u \ dV = \int_V T_{ij}D_{ij}\ dV - \int_S \mathbf{q} \cdot \mathbf{\hat{n}} \ dS + \int_V \rho h \ dV $$ (4.9)
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 *

 * }

From the divergence theorem (equation 1.4) the surface integral can be expressed as a volume integral. The remaining volume integrals can be expressed for an arbitrary volume V within the continuous medium. Then the conservation of energy equation takes the form below.

 * {| style="width:100%" border="0"

$$\displaystyle \rho \frac{du}{dt} = T_{ij}D_{ij} - \frac{\partial q_j}{\partial x_j}  +  \rho h $$ (4.10)
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 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Description
When the displacements and displacement gradients are small, the second Piola-Kirchoff stress tensor $$\displaystyle \mathbf{\tilde{T}}$$ is approximately equal to the Cauchy stress tensor $$\displaystyle \mathbf{T}$$. Thus for small vibrations of an elastic system the balance of linear momentum equations, in terms of $$\displaystyle \mathbf{\tilde{T}}$$, reduces to

 * {| style="width:100%" border="0"

$$\displaystyle \frac{\partial T_{ji}}{\partial X_j} + \rho_0 b_{0i} = \rho_0 \frac{\partial^2 u_i}{\partial t^2} $$     (5.1)
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 * style="width:10%; |
 *

 * }

Consider Hooke's Law $$\displaystyle \left( T_{ij} = \lambda e \delta_{ij} + 2G \epsilon_{ij} \right)$$ where $$\displaystyle e=\epsilon_{kk}$$ and $$\displaystyle \lambda$$ and $$\displaystyle G $$ are the Lame elastic moduli.

 * Show that the first equation of motion then leads to the following displacement equation of motion: $$\displaystyle (\lambda + G) \frac{\partial e}{\partial X} + G \nabla^2 u_x + \rho_0 b_{0x}=\rho_0 \frac{\partial^2 u_x}{\partial t^2}$$.
 * Write the vector equation of which is the first component. ← (I'm not sure what this question is asking.)

Solution
First substitute Hooke's Law for the stress tensor into equation 5.1 to get

 * {| style="width:100%" border="0"

$$\displaystyle \lambda \frac{\partial}{\partial X_j} \left( e \delta_{ij} \right ) + 2G \frac{\partial \epsilon_{ij}}{\partial X_j} + \rho_0 b_{0i}=\rho_0 \frac{\partial^2 u_i}{\partial t^2} $$.     (5.2)
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 *

 * }

The strain tensor is defined as

 * {| style="width:100%" border="0"

$$\displaystyle \epsilon_{ij} = \frac{1}{2} \left( \frac{\partial u_i}{\partial X_j} + \frac{\partial u_j}{\partial X_i} \right) $$.     (5.3)
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 * style="width:10%; |
 *

 * }

Substituting this into equation 5.2 yields

 * {| style="width:100%" border="0"

$$\displaystyle \lambda \frac{\partial}{\partial X_j} \left( e \delta_{ij} \right ) + G \frac{\partial }{\partial X_j} \left(\frac{\partial u_i}{\partial X_j} + \frac{\partial u_j}{\partial X_i} \right ) + \rho_0 b_{0i}=\rho_0 \frac{\partial^2 u_i}{\partial t^2} $$.     (5.4)
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 * style="width:10%; |
 *

 * }

The Laplacian of a vector $$\displaystyle u_x$$ is defined as

 * {| style="width:100%" border="0"

$$\displaystyle \nabla^2 u_x = \frac{\partial^2 u_x}{\partial x^2} + \frac{\partial^2 u_x}{\partial y^2} +\frac{\partial^2 u_x}{\partial z^2} $$.     (5.5)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Substitution of this into 5.4 yields

 * {| style="width:100%" border="0"

$$\displaystyle \lambda \frac{\partial e}{\partial X} + G \left( \nabla^2 u_x + \frac{\partial e}{\partial X}\right ) + \rho_0 b_{0x}=\rho_0 \frac{\partial^2 u_x}{\partial t^2} $$.     (5.6)
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 * style="width:10%; |
 *

 * }

Simplifying results in the displacement equation of motion.

 * {| style="width:100%" border="0"

$$\displaystyle (\lambda + G) \frac{\partial e}{\partial X} + G \nabla^2 u_x + \rho_0 b_{0x}=\rho_0 \frac{\partial^2 u_x}{\partial t^2} $$     (5.7)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Description
In an ideal frictionless fluid there are no shear tractions on any plane.
 * 1) Show that his implies that $$\displaystyle T_{ij}=-p \delta_{ij}$$.
 * 2) Show that for this case the stress power, $$\displaystyle T_{ij}D_{ij}$$, is given by

 * {| style="width:100%" border="0"

$$\displaystyle T_{ij}D_{ij}=\frac{p}{\rho} \frac{d \rho}{dt} $$     (6.1)
 * style="width:10%; |
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 *

 * }

 * Express this result in terms of the specific volume $$\displaystyle \nu = \frac{1}{\rho}$$.

Solution
The stress vector/tensor relationship is defined by $$\displaystyle t_i^{(n)} = T_{ji}n_j$$ where the superscript n corresponds to the number of equations or rows of the stress tensor. Since $$\displaystyle T_{ji}$$ is a second order tensor with 9 components, $$\displaystyle n=1,2,3$$.

 * No shear tractions implies that $$\displaystyle T_{ji}=0 $$ when $$\displaystyle i \neq j$$.
 * For hydrostatic stress, the static pressure is defined as $$\displaystyle -p= \frac{1}{3} T_{ii}$$. Therefore, $$\displaystyle T_{ji}=-p $$ when $$\displaystyle i=j$$. This can also be written using the Kronecker delta as

 * {| style="width:100%" border="0"

$$\displaystyle T_{ij}=-p \delta_{ij} $$     (6.2)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *

 * }

The definition of the rate of deformation tensor is $$\displaystyle D_{ij} = \frac{1}{2} \left( v_{i,j} + v_{j,i} \right)$$. Since $$\displaystyle \mathbf{T}$$ is symmetric, $$\displaystyle D_{ij} = v_{i,j} $$. So from equation 6.2 it can be noted that

 * {| style="width:100%" border="0"

$$\displaystyle T_{ij}D_{ij}=-p \delta_{ij} \ v_{i,j} \qquad \Rightarrow \qquad T_{ij}D_{ij}=-p v_{i,i} \qquad \text{when} \quad i=j $$.     (6.3)
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 * style="width:10%; |
 *

 * }

Recognizing that $$\displaystyle v_{i,i}$$ is the time derivative of the displacement vector,

 * {| style="width:100%" border="0"

$$\displaystyle T_{ij}D_{ij}=-p \frac{d}{dt} \left(u_{i,i} \right) $$     (6.4)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

 * which can be expressed alternatively below by recalling that $$\displaystyle v_{i,i} = \frac{\dot{\rho}}{\rho}$$
 * {| style="width:100%" border="0"

$$\displaystyle T_{ij}D_{ij}=-p \left( \frac{1}{\rho} \frac{d \rho}{dt} \right) \quad = \quad \frac{-p}{\rho} \frac{d \rho}{dt} $$     (6.5)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 *

 * }

Description
Show that in the frictionless fluid of Problem 6, if there is no internal heat source and if the heat conduction is governed by Fourier's Law $$\displaystyle \mathbf{q}=-\kappa \nabla \theta $$ where $$\displaystyle \theta $$ is the temperature, then the energy becomes

 * {| style="width:100%" border="0"

$$\displaystyle \rho \frac{du}{dt} = \nabla \cdot \left( \kappa \nabla \theta \right) - p \nabla \cdot \mathbf{v} $$     (7.1)
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 *

 * }

Solution
Substituting equation 6.3 into 4.10 when $$\displaystyle h=0 $$ returns

 * {| style="width:100%" border="0"

$$\displaystyle \rho \frac{du}{dt} = - p v_{i,i} - \frac{\partial q_j}{\partial x_j} $$.     (7.2)
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 *

 * }

Inputting the expression for Fourier's Law and recognizing that the first expression on the right-hand side is the divergence of the velocity vector

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$$\displaystyle \rho \frac{du}{dt} = - p \nabla \cdot \mathbf{v} + \frac{\partial}{\partial x_j} \left( \kappa \nabla \theta \right) $$.     (7.3)
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 *

 * }

Now the second expression on the right-hand side is the divergence, so equation 7.3 can be rewritten and rearranged to obtain the conservation of energy expression in a frictionless fluid.

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$$\displaystyle \rho \frac{du}{dt} = \nabla \cdot \left( \kappa \nabla \theta \right) - p \nabla \cdot \mathbf{v} $$     (7.4)
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 *

 * }

Description
By using the energy equation show that the Clausius-Duhem inequality takes the form

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$$\displaystyle \rho \left( \theta \dot{s} - \dot{u} \right) + \mathbf{T}:\mathbf{D} - \frac{1}{\theta} \mathbf{q} \cdot \nabla \theta \ge 0 $$.     (8.1)
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 * }

Solution
From the equation of energy (4.10),

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$$\displaystyle \rho \dot{u} = \mathbf{T}:\mathbf{D} - \nabla \cdot \mathbf{q} +  \rho h $$ (8.2)
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 *

 * }

Based on the second law of thermodynamics, the Clausius-Duhem inequality states that the rate of entropy increase is greater than or equal to the entropy input rate. In other words

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$$\displaystyle \dot{s} - \frac{h}{\theta} + \frac{1}{\rho \theta} \nabla \cdot \mathbf{q} - \frac{\mathbf{q}}{\rho \theta^2} \cdot \nabla \theta \ge 0 $$.     (8.3)
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 * }

Multiplying through by $$\displaystyle \rho \theta$$ yields

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$$\displaystyle \rho \theta \dot{s} - \rho h + \nabla \cdot \mathbf{q} - \frac{\mathbf{q}}{\theta} \cdot \nabla \theta \ge 0 $$.     (8.4)
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 * }

Solving equation 8.2 for $$\displaystyle \rho h $$ and substituting into 8.4

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$$\displaystyle \rho \theta \dot{s} - \left( \rho \dot{u} - \mathbf{T}:\mathbf{D} + \nabla \cdot \mathbf{q} \right) + \nabla \cdot \mathbf{q} - \frac{\mathbf{q}}{\theta} \cdot \nabla \theta \ge 0 $$.     (8.5)
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 *

 * }

The two divergence terms cancel out, and the remaining terms can be collected to express the Clausius-Duhem inequality as requested.

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$$\displaystyle \rho \left( \theta \dot{s} - \dot{u} \right) + \mathbf{T}:\mathbf{D} - \frac{1}{\theta} \mathbf{q} \cdot \nabla \theta \ge 0 $$     (8.6)
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 * style="width:10%; padding:10px; border:2px solid #8888aa" |
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 * }

Description
As an example of how the Clausius-Duhem inequality palces restrictions on possible constitutive equations, consider a class of viscous fluid assumed defined by four constitutive equations of the forms below in terms of entropy $$\displaystyle s$$, density $$\displaystyle \rho$$, and rate of deformation $$\displaystyle \mathbf{D}$$.

 * Stress $$\displaystyle \Rightarrow \quad \mathbf{T} = \mathbf{T} (s,\frac{1}{\rho},\mathbf{D})$$
 * Heat flux vector $$\displaystyle \Rightarrow \quad \mathbf{q} = \mathbf{q} (s,\frac{1}{\rho},\mathbf{D})$$
 * Internal energy $$\displaystyle \Rightarrow \quad u = u (s,\frac{1}{\rho},\mathbf{D})$$
 * Temperature $$\displaystyle \Rightarrow \quad \theta = \theta (s,\frac{1}{\rho},\mathbf{D})$$

Show that the result of Problem 8 takes the form

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$$\displaystyle \rho \left( \theta -\frac{\partial u}{\partial s}\right) \dot{s} -\rho \frac{\partial u}{\partial D_{km}}\dot{D_{km}} + \frac{\partial u}{\partial(1/ \rho)} \frac{\dot{\rho}}{\rho}+ T_{km}D_{km} - \frac{1}{\theta} q_k \frac{\partial \theta}{\partial x_k} \ge 0 $$.     (9.1)
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 * }

If this is to hold for arbitrary independent prescriptions of $$\displaystyle \dot{s}$$ and $$\displaystyle \dot{D_{km}}$$ we must have $$\displaystyle \theta=\partial u / \partial s $$ and $$\displaystyle \partial u / \partial D_{km}=0$$. Hence $$\displaystyle u$$ is independent of $$\displaystyle \mathbf{D}$$. If we define the thermodynamic pressure by $$\displaystyle p=-\frac{\partial u}{\partial (1/\rho)}$$ and use the equation of continuity, show that the inequality then reduces to

 * {| style="width:100%" border="0"

$$\displaystyle (T_{km} + p \delta_{km})D_{km} - \frac{1}{\theta}q_k \frac{\partial \theta}{\partial x_k} \ge 0 $$.     (9.2)
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 * }

Solution
Equation 8.6 can be rewritten in Cartesian coordinates as

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$$\displaystyle \rho \theta \dot{s} - \rho \dot{u} + T_{km}D_{km} - \frac{1}{\theta} q_k \cdot \frac{\partial \theta}{\partial x_k} \ge 0 $$     (9.3)
 * style="width:10%; |
 * style="width:10%; |
 *

 * }

Description
If $$\displaystyle \sigma_{ij} = -p \delta_{ij}$$, determine an equation for the rate of change of specific entropy during a reversible thermodynamic process.


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User:Eml5526.s11.team7.jb/prob66

Problem Description
Solve the general 2-D model with data set 1 (G2DM1.0/D1) using 2-D Linear Lagrangian element basis function (LLEBF) by increasing the number of elements until an accuracy of $$\displaystyle 10^{-6}$$ is reached at the center $$\displaystyle (x,y)=(0,0)$$.

Integrate $$\displaystyle \mathbf{k^{e}}$$ in parent coordinates $$\left \{ \xi_{i} \right \}$$. Find $$\displaystyle \mu$$ to integrate $$\displaystyle \mathbf{k^{e}}$$ exactly with Gauss-Legendre quadrature (GLQ)

Given
For G2DM1.0/D1, the governing partial differential equation (PDE) and given conditions are,

 * $$\displaystyle \frac{\partial}{\partial x_{i}} \left[ \boldsymbol{\kappa_{ij}} \frac{\partial u}{\partial x_{j}} \right ] + f = \rho c \frac{\partial u}{\partial t} $$

 * where
 * The domain $$\displaystyle \Omega = \overline{\omega}$$ is a biunit square

 * $$\displaystyle \boldsymbol{\kappa_{ij}}=\boldsymbol{I}$$

 * $$\displaystyle f = 0$$

 * $$\displaystyle \bar{m} \frac{\partial u}{\partial t } = 0 $$ (static case)

So the PDE simplifies to:
 * $$\displaystyle \frac{\partial}{\partial x_{i}} \left[ \boldsymbol{I} \frac{\partial u}{\partial x_{j}} \right ] = 0$$

The natural boundary condition is not prescribed and the essential boundary condition is defined below:

 * $$\displaystyle g=2$$ on $$\displaystyle \partial u$$

Generating LLEBF
The 2-d Lagrangian interpolation function can be formed as follows
 * $$\displaystyle N_{I}^{e}(x,y)=L_{i,m}^{e}(x)\cdot L_{i,m}^{e}(y)$$

where
 * $$\displaystyle I=i+(j-1)m$$

For linear basis functions there are two nodes per element in along each dimension $$\displaystyle n=m=2$$. So for each element there are four nodes and four corresponding basis functions.
 * $$\displaystyle N_{1}^{e}(x,y)=L_{1,2}^{e}(x)\cdot L_{1,2}^{e}(y)=\frac{(x-x_{2})(y-y_{2})}{(x_{1}-x_{2})(y_{1}-y_{2})}$$
 * $$\displaystyle N_{2}^{e}(x,y)=L_{2,2}^{e}(x)\cdot L_{1,2}^{e}(y)=\frac{(x-x_{1})(y-y_{2})}{(x_{2}-x_{1})(y_{1}-y_{2})}$$
 * $$\displaystyle N_{3}^{e}(x,y)=L_{1,2}^{e}(x)\cdot L_{2,2}^{e}(y)=\frac{(x-x_{2})(y-y_{1})}{(x_{1}-x_{2})(y_{2}-y_{1})}$$
 * $$\displaystyle N_{4}^{e}(x,y)=L_{2,2}^{e}(x)\cdot L_{2,2}^{e}(y)=\frac{(x-x_{1})(y-y_{1})}{(x_{2}-x_{1})(y_{2}-y_{1})}$$

For example if one element is used the basis function for the first element on that element is as follows:
 * $$\displaystyle N_{1}^{1}(x,y)=\frac{(x-1)(y-1)}{(-1-1)(-1-1)}$$

One can see from the plot of this basis function that it is equal to one at (-1,-1) and equal to zero at all other nodal points and thereby satisfies the CBS:

The following matlab code was adapted from Problem 6.5 to solve the (G2DM1.0/D1) using 2-D LLEBF.

The exact solution is found simply by analyzing the data given in the problem. One can see the temperature field does not change with time as given by $$\displaystyle \frac = 0$$. There is no internal heat generation as given by $$\displaystyle f=0$$. Lastly the conductance matrix is equal to the identity matrix meaning that the material is isotropic. The temperature is given around the entire domain as 2 (essential boundary condition) and no flux is applied (natural boundary condition). So the biunit square is isotropic, has temperature two around its entire border with no flux, internal heat generation, or change in temperature. The only possible conclusion is that the exact solution is $$\displaystyle {u(x,y)=2}$$.

So the temperature at the center $${u(0,0)=2}$$. The approximate solution yielded $${u^h(0,0)=2}$$. We find the error to be zero using only one element.

 * $$\displaystyle \tilde{k}(w,u)=\int_{\Omega^{}}^{.} \nabla w \boldsymbol{\kappa}\nabla u d\Omega$$
 * $$\displaystyle \tilde{m}(w,u)=\int_{\Omega^{e}}^{.} w\rho c\frac{\partial u}{\partial t}d\Omega=0$$
 * $$\displaystyle \tilde{f}(w)=-\int_{\Gamma_{h}^{e}} whd\Gamma_{h}^{e}+\int_{\Omega}  wfd\Omega=0$$

 * $$\displaystyle \tilde{k}_{ij}^{e}=\int_{\omega^{e}}^{.} \nabla(N_{I}^{e}) \boldsymbol{\kappa}\nabla(N_{J}^{e})d\omega^{e}$$

 * $$\displaystyle \tilde{k}_{ij}^{e}=\int_{\omega^{e}}^{.} \left [\boldsymbol{J}^{e*-T}\nabla_\xi N_{I}^{e}(\xi))\right ] \boldsymbol{\kappa}(\xi) \left [\boldsymbol{J}^{e*-T}\nabla_\xi N_{J}^{e}(\xi))   \right ] |\boldsymbol{J}| d\omega^{e}$$

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User:Egm6341.s11.team1.arm/HW6

Problem 7: Methods of computing cicumferences
 Solved without assistance

Given: Bifolium and ellipse
Bifolium

Ellipse

Arc length

Ellipse circumference

Find: Circumference using different methods
Compute the circumference of a bifolium and ellipse using the following methods: a) Composite trapezoidal rule b) Romberg's method c) Clenshaw-Curtis quadrature (Trefethen's code) d) Clenshaw-Curtis quadrature (Winckel's code) Compute the circumference of the ellipse using both the arc length method and the elliptic integral method.

Composite Trapezoidal Method
The MATLAB code below was developed to calculate the circumference of the bifolium and the given ellipse. The circumference of the ellipse was calculated two ways: the arc length method (Eq. 7.3), and the elliptic integral method (Eq. 7.4).

Romberg's method
The MATLAB code below was developed to calculate the circumference of the bifolium and the given ellipse. The circumference of the ellipse was calculated two ways: the arc length method (Eq. 7.3), and the elliptic integral method (Eq. 7.4).

bifolium
solution here

ellipse, where $$ a=10 \ $$ and $$ b=3 \ $$ using $$ r( \theta\ )=\frac{a(1-e^2)}{1-e \cos \theta\ } \ $$ and $$ C = \int_{ \theta\ =0}^{2 \pi\ } dl \ $$
solution here

ellipse where $$ a=10 \ $$ and $$ b=3 \ $$ using $$ E(e) = \int_{0}^{\frac{ \pi\ }{2}} \left [ 1-e^2 \sin \theta\ \right ] ^{\frac{1}{2}}, d \theta\ \ \ $$ and $$ C = 4AE(e) \ $$
solution here

bifolium
solution here

ellipse, where $$ a=10 \ $$ and $$ b=3 \ $$ using $$ r( \theta\ )=\frac{a(1-e^2)}{1-e \cos \theta\ } \ $$ and $$ C = \int_{ \theta\ =0}^{2 \pi\ } dl \ $$
solution here

ellipse where $$ a=10 \ $$ and $$ b=3 \ $$ using $$ E(e) = \int_{0}^{\frac{ \pi\ }{2}} \left [ 1-e^2 \sin \theta\ \right ] ^{\frac{1}{2}}, d \theta\ \ \ $$ and $$ C = 4AE(e) \ $$
solution here

title: User:Egm6341.s11.team1.arm/HW6


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User:EML5526.S11.Team1.HS/HW6

=HW6=

Given
6.1)

Given vector $$ q_x = -y^2 ~,~ q_y = -2xy $$ on domain shown in Figure 6.2.

6.2)

Given a vector field $$ q_x = 3x^2 y +y^3 ~,~ q_y = 3x + y^3 $$ on the domain shown in Figure 6.9. The curved boundary of the domain is a parabola.

6.3) Given $$ \oint_{T} \mathbf{n} d\Gamma = 0 $$

Find
6.1)

Verify the divergence theorem.

6.2)

Verify the divergence theorem.

6.3)

Prove using divergence theorem

FB 6.1)

We have to show that:

Where $$ \Omega $$ is the domain, $$ \Gamma $$ is the boundary, n is the normal vector and the notation $$ q_{i,i} $$  means $$ \frac{ \partial q_i}{ \partial x_i} $$

The integration on the boundary is

FB 6.2)
To calculate the integral on the surface we see that part of the surface is the parabola and the other part is in the x axis; therefore the integral will be:

The first integral is on the parabola and the second part is on the x axis.

On the curved surface the, normal vector on a segment of surface ds is shown in the figure. From that figure we observe that:

Also, it is clear that on the x axis we have y=0. Replacing in the integral we obtain:

Taking the derivative on the equation of the parabola we obtain:

Replacing on the integral we have:

Changing the limits of integration in the first two terms and organizing we obtain:

The integral on the volume will be:

Performing first the integration with respect to y we have:

FB 6.3)
The theorem says that for any tensor $$ T_{i...j} $$ (of any order) inside a volume $$ \Omega $$ surrounded by a surface $$ \Gamma $$ with a normal vector n

In this case the tensor T is the constant 1; therefore, changing the integral of surface to an integral of volume we have

But as 1 is a constant, its derivatives with respect to any coordinate are zero and we have

=Problem 6.5: Solve G2DM10/D1 using 2DLIBF=

Given: 2DLIBF from p. 29-2
PDE

$$ K = I $$ (identity matrix)

$$ f = 0, \frac{\partial u}{\partial t} = 0 $$

$$ \frac{\partial}{\partial x_i} \left(k_{ij} \frac{\partial u}{\partial x_j} \right) = 0 $$

Essential B.C.

$$ g = 2 on  \partial \Omega $$

Natural B.C.

none

Find
Solve G2DM10/D1 for $$ u^h $$ until accuracy $$ 10^{-6} $$ at center $$ (x,y) = (0,0) $$

Solution
Using the weak form to be

and

2DLIBF

where

$$ I = i + \left( j-1 \right) m $$

$$ n = m = 2, 4, 6, 8... $$

Thus

Therefore by substituting equation 6.5.3 into equation 6.5.1 and integrating, the stiffness matrix can be solved for. This is done by using Matlab.

The solution yields

title: User:EML5526.S11.Team1.HS/HW6


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User talk:Anand ST14

=Problem 6.3=

Given
Calculix Tutorials by Beede in four parts posted on Mechanical Hacks Blog as per the links provided below : -Calculix FEA Beam Part 1: Buliding Geometry and Meshing -Calculix FEA Beam Part 2: Exporting Mesh, Loads, and Boundary Conditions -Calculix FEA Beam Part 3: Writing an Input File for the CCX Solver -Calculix FEA Beam Part 4: Post Processing Results in CGX

Find
To reproduce the steps for modeling and analysis of a beam problem from above mentioned Tutorials through CGX and CCX Finite Element software

Start up data
$$ 1.$$ open the SciTE text editor, then “save as” a blank file with name beam.fbd. Firing up cgx with the current file as input is available from the Tools->Pre Process menu item or the F10 key. Details are shown as per Figure 6.3.1(1), Figure 6.3.2(1) and Figure 6.3.3 (1)

$$ 2.$$It is easier to define geometry with a batch file. Interactive mode is a bit more useful when it comes to working with the mesh like assigning boundary conditions, loads, and constraints to a set of specific nodes.

Add the following text to the beam.fbd file as shown in the below Figure 6.3.4 (1).

The first two lines of the batch file define points in cgx model space. The first point is named p1 located at (x,y,z) = 0. The third line of code defines a cgx line with name l1 that starts at point p1 and ends a point p2 and has a division of 25.

Hit the F10 Key or choose Tools->Pre Process from the menu.

The following result is displayed in the below Figure 6.3.6 (1).

In addition the window is divided into two distinct areas. On the right side top corner is the graphics display area bordered by a thin black line. Inside the graphics display area clicking and holding the mouse has the following functionality :

1. Left mouse button = rotate 2. Middle mouse button = zoom 3. Right mouse button = pan

Moving the mouse out of the area surrounded by the thin black border the mouse has different functionality. In this area the left mouse button is used to bring up a menu which gives the user a graphical front end to some of the cgx commands which can alternatively be typed on the keyboard. Only a small subset of the commands are available through the pop up menu.

A simple 100 x 10 x 1 rectangular beam
We now proceed to generate a simple 100 x 10 x 1 rectangular beam meshed with 1x1x1 cubes. We revise the generated batch file $$ beam.fbd $$ with more points and lines as shown in the following Figure 6.3.7 (1) and Figure 6.3.8 (1).

These extra segments will allow for meshing at the desired resolution of 1x1x1 size elements. The reason for this is that each line segment is internally limited to 99 divisions. We also note that quadratic elements mesh at a resolution of 2 line divisions equals 1 mesh element side length. Linear elements mesh at a rate of 1 division equals 1 mesh element length.

The output window is capturing all the cgx program output and keyboard input. While the cgx window is the active window any keyboard input will be shown on the bottom line of the output window as it can be seen that two line input is provided in above figure 6.3.9 (1) - -plot pa all tells cgx to display all points p with their names a. More specifically it displays all the points contained within the cgx set named all. Users can define their own sets with arbitrary members to manage the problem definition and processing. -The second command plus la all is used to add the lines with their names to the existing plot. If instead plot la all had been typed after the first command the points and their names would have been eliminated from the new plot view.

Building a surface from the line using a sweep command
We now proceed to build a surface from the line using a sweep command. We revise the generated batch file $$ beam.fbd $$ as shown in the following Figure 6.3.10 (1).

The text file is updated to create a set containing the four lines and then to sweep the set in the y direction for a length of 10 units. The command seta is used to create a new set named lines. The l flag indicates that the items to add are lines. Several seta commands can be used to incrementally add elements, nodes, or geometry to the same set. The swep command sweeps the set named lines via translation in (0,10,0) with 10 divisions and stores the newly created lines and points into a set labeled sweplines. The plus sa all command shows all surfaces with their labels.

Use of plot ld all command
We use the plot ld all command to show only the lines with their divisions. The divisions in Figure 6.3.11(1) define the mesh resolution of 100 x 10 in the x and y- directions respectively.

The following operation as shown in Figure 6.3.12-1(1) adds another swep to turn the 2D surface into a 3D body. The .fbd file and results are shown.

In figure 6.3.12-2 (1) the menu system toggles the background color as explained in the middle of above Figure 6.3.12-1(1) -

Element types available to mesh with using cgx
The following element types are available to mesh with using cgx -

1. be2: 2 node 1D linear beam element 2. be3: 3 node 1D quadratic beam element 3. tr3: 3 node 2D linear triangular shell element 4. tr6: 6 node 2D quadratic triangular shell element 5. qu4: 4 node 2D linear quadrangle element 6. qu8: 8 node 2D quadratic quadrangle element 7. he8: 8 node 3D linear brick element 8. he20: 20 node 3D quadratic brick element 9. pe6: 6 node 3D linear penta element 10. pe15: 15 node 3D quadratic penta element

Meshing using interactive mode of CGX
We load the beam defined in the previous step in cgx and keep the mouse must be within the cgx window. Then provide with following commands

The elty all he8 command indicates that we will be meshing the set all with he8 elements.

Various views of the meshed model
Different graphical views of the model can be generated as shown in below Figure 6.3.14(1) :

1. Viewing -> Toggle Model Edges in the area outside of the graphics display part of cgx : as shown in Figure 6.3.14(1) below -

2. Viewing->DOTS and Viewing->Toggle Element Edges :as shown in Figure 6.3.15(1) below -

In this section, the mesh which was created based on the defined geometry and selected element type.

Creation of Mesh-File
Now we define into the batch file beam.fbd, a set named beam, mesh it, and save the nodes and elements in Abaqus format for input into ccx. The following Figure 6.3.1 (2) shows the beam.fbd file after adding the new commands

On line 1 the command seto beam has been used. This command tells cgx to open a set named beam. While beam is the open set geometry created using cgx commands will be added to this set. Finally the set is closed on line 16 using the setc beam command. Lines 16 and 17 define the element type and build the mesh.

Now that the mesh has been created the command send beam abq is used to export the mesh created from set beam to the abq (Abaqus) format. The result of this command will reside in the same directory as the open .fbd file that created it, the working directory. Its name will be beam.msh.

Next hit the F10 key or Tools->Pre Process to process the batch file. The following window as given in Figure 6.3.2(2) appears and the mesh file is written to disk.

Contents of the Mesh-file
The most interesting parts are the start of the node and element definitions respectively as shown in below figure 6.3.3(2) and Figure 6.3.4(2) - These commands capture the shape and number ordering of the mesh.

Line number 1 above starts with the command *NODE, NSET=Nbeam which declares the start of a node list. It also assigns a name to this set of nodes, Nbeam. Grouping together nodes and elements into logical units is a convenience of management.

In the second image the declaration for the elements is observed on line number 2224. This command *ELEMENT, TYPE=C3D8, ELSET=Ebeam defines linear 3D bricks with 8 nodes and groups them in the Ebeam set. In cgx he8 elements turn into C3D8 elements within ccx.

Note - To define nodes and elements for a low resolution mesh the same commands shown in the images above can be written by hand in an .inp file.

To export some loads and boundary conditions
This will require selecting some nodes within interactive mode to create a set. Then the set will be exported using the send command.

With the beam.fbd file open in cgx issue the following commands to view only the nodes of the mesh. The commands are shown at the bottom of the Figure 6.3.4 (2) above.

The command plot n all is used to display the nodes, n, from the set all. The results should look like the image below. The rot -z command rotates the view to the -z direction. In this case the set all contains everything defined in the cgx file, the same result could also be obtained by using the command plot n beam because beam is the only set meshed thus far.

Graphical selection
We use the command qadd fixed and press the enter button so that the mouse cursor has changed to look like Figure 6.3.7(2) below -

Variable size Selection of the graphical area can be done using command - qadd and r-button on the keyboard. This is shown in below Figure 6.3.8(2)

In order to select all the nodes of the left hand edge the selection area will be made large enough to encompass the entire edge. In addition there are two selection modes used by the qadd command. The default selection mode adds one item each time the corresponding keyboard button is pressed. This works good for selecting one of a particular item with a small selection rectangle. If the desire is to bulk add all the items located within the selection rectangle the mode:a should be used. While the qadd command is active press the a button on the keyboard to active bulk add. The following Figure 6.3.9(2) and Figure 6.3.10(2) show the larger selection rectangle and the results of pressing the a button to active mode:a. The results of this selection will be shown in the output window as a numbered list of nodes which were added to the set. This is shown at the bottom of the figures below.

Now that the nodes have been added to the set fixed the qadd command can be quit using the q button. In order to visualize the nodes contained in the set fixed the view can be manipulated to look like the following Figure 6.3.11(2) and Figure 6.3.11a(2). Changing the color by adding a color character at the end of the plus or plot command is helpful to confirm the set contains the desired items.

The command view edge off is equivalent to Viewing->Toggle Model Edges. Next the plus l beam w command adds the lines from the set beam to the plot in white. The last command issued adds the nodes contained in the new set fixed. These are plotted in green using the plus n fixed g command.

Boundary Conditions
Now that the nodes for the fixed end of the beam have been grouped together they can be exported with a boundary condition. In this case each node will be constrained such that displacements (x,y,z) = 0, that is they are fixed. This is accomplished using the send command shown in the Figure 6.3.12(2) below.

The command send fixed abq spc 123 is used to write the boundary condition to a file. In this case fixed is the name of the set of nodes, abq defines the Abaqus format, spc stands for single point constraint, and 123 indicates the degrees of freedom to fix.

We notice the output below the send command. The boundary conditions are written to a file named fixed_123.bou in the working directory. Part of it is shown in figure 6.3.13(2) below. Also we notice the output statement ready which indicates processing of the file has finished.

Loads
We define a distributed pressure load on an element face. Use the plot f beam command to view the element faces as shown in Figure 6.3.14(2).

Next the send command is used to write a pressure load to the load.dlo file. The send command is used to output a pressure, pres, with magnitude equal to 10 as follows -

The resulting load.dlo file has the following contents within -

Line number 2 defines a load of magnitude 10 applied to the 6th face, P6, of element number 751. To confirm the correct element has been selected examine the model using the plot ea beam command. Also, note the Viewing->Toggle Culling Back/Front mode is used to see inside of the elements. This is necessary because the element labels are hidden behind the outside faces.

In order to confirm the face selected in the previous step is indeed number 6 it will be necessary to turn on node number labels and zoom in on the area of element 751.

Following Figure 6.3.18(2) shows the order of the nodes for element 751.

We compare this to the element numbering guide shown in the Calculix user manual. Node numbers 1673, 1674, 1675, 1676 are located on the right hand face at x=100. These node numbers correspond to 1, 2, 3, 4 in the figure below. This figure has been copied directly from the Calculix manual. It also contains details for different element shapes. Thus it is clear to see face 6 consists of nodes 1673, 1676, 1677, 1680 and is located exactly where it should be.

The other loading method mentioned above occurs at a specific node, point loading. Loading at a point contrasts with loading a face. When a point is loaded a force vector consisting of three components for each of the x,y,z directions is applied to it. When a pressure is defined it is considered to act normal to the element face and is distributed across the surface area. If the orientation of the face or its area change the corresponding direction and magnitude of the resulting force will update as well. In the case of point loading the force will be constant in magnitude and direction as the object changes shape. Use the following commands to define a new set with the two nodes at the upper right corner of the beam where x=100, y=10 and z=0 or z=1. These nodes are a member of element 751 shown above.

The first command adds the nodes to a new set. This is followed by a send command to output the force in the Abaqus format with the vector components (0,-5,0). The resulting loadnodes.frc file contents are shown in the following image. Each node of the set is listed with the direction of the applied force and followed by the magnitude.

Now the mesh, boundary condition, and loading has been defined and exported to files and it will be necessary to write a Calculix solver (ccx) input file (.inp) which includes these exported files to define the problem.

Input file
Here we start by writing an input file and solving for the results using the finite element method. The input file describes the finite element problem as an ordered set of commands with parameters. We Open SciTE and create a new file named beam.inp. This file is located with the files from the previous articles. Then we add the following code from the figure 6.3.1(3) below to this file.

The first two lines of the input file declare a *HEADING where the user can write comments and notes on the following line. The next command, *INCLUDE, inserts the contents of the file named beam.msh into the input file. The next command, *INCLUDE, inserts the contents of the file named beam.msh into the input file. On lines 6 thru 9 the material model is defined.

Solid Section Definition
This assigns the elastic material EL to the element set Ebeam.

Loading Step
In this step the loading is applied using the *STATIC method with only one step used. This indicates the force will be slowly applied in a quasi-static sense. This will neglect the mass inertia of the structure. a distributed load is applied to the tip of the beam. *NODE PRINT and *EL PRINT commands are used to print the results to a .dat text file.

Output
With the beam.inp file ,we open in SciTE choose Tools->Solve or we use the keyboard shortcut Ctrl+F10 to process this file with ccx. This will solve the finite element model as defined and output the requested files.

The output of the ccx program appears in the SciTE output window. It provides detailed diagnostics information. At the bottom it finishes the job and exits with code 0. This successful indication means a solution was found and the results have been written to disk. The ccx output should look like the following Figure 6.3.3(3) and Figure 6.3.4(3). Progress can be monitored by watching the output as ccx attempts to find a convergent solution or steps through a number of loading conditions.

Von-Mises stress distribution
Using the cgx menu system click on Datasets -> 2 STRESS  1.000000, we use the menu system again to click on Datasets -> Entity -> Mises. This displays the Von Mises stress distribution for the solution at step 1 in figure 6.3.5(3) below.

Calculix FEA Beam Part 4: Post Processing Results in CGX
This article will explore some of that functionality by examining the results of the cantilever beam problem from the previous article.

Open the input file beam.inp in SciTE. Choose Tools -> Post Process from the menu system. This launches cgx with the beam.frd results file.

Note-Menu option is only available when working with an .frb file containing some simulation results e.g. Stress, Displacement etc, as seen from above figure 6.3.1(4).

plotting for various entities
The stresses and the displacement are shown in contour or graph format as below in figure 6.3.2 (4) and figure 6.3.3 (4) An option from within the -Entity- sub-menu results in a colored plot of the selected field information.

Plot magnification
To scale up the displacement use the keyboard command scal d 10000. This will scale the deformed shape by a factor of 10000. This is visible in Figure 6.3.3(4) below-

We toggle the displacement to visualize the deformed shape. The resulting displacement from this simulation is quite small. It is not noticeably visible.

The viewing mode can be returned to the default gray beam by choosing Viewing->Show All Elements With Light as shown in below figure 6.3.4(4)

Animation
Another option to visualize the shape of the deformed structure is through animation as shown in below Figure 6.3.6(4)

> Options within the animation menu are as follows -

>

To specify a cut plane
The cut plane is defined by three nodes. The nodes can be specified from the menu system using the Cut menu.

> Result of cutting the beam top to bottom is getting the top of the beam is in tension and the bottom in compression as seen from the below figure 6.3.9(4).

>

Total displacement at nodes
As shown in Figure 6.3.10(4), we plot the total displacement of nodes along the top of the beam at y=10 and z=0. They span along the x-axis starting at the fixed end and progressing towards the free end. The desired field for plotting must be the currently active field selected from Datasets.

>

Finally, the following graph is promptly displayed for displacement values at various nodes.

>

Nodal values of entities
The nodal values are printed in yellow using the command plus nv all y. The following Figure 6.3.12(4) shows top side of the beam’s free end, near x=100, y=0, z=0.

>

Reduced set of nodes
Viewing the entire beam is difficult when several thousand nodal values are print on the screen. Limiting the nodal values to an interesting set is likely the best use of this tool.

>

This completes the 4 steps of beam analysis using CGX and CCX of calculix software

=Problem 6.4-1=

Given
Referring to page 2 of for problem assignment. This slide refers to Fish and Belytschko Problem 6.1 on page 148. The vector field is given as follows -

$$\displaystyle (Eq. 6.4.1-1) $$ Also the given domain is a square as shown in below Figure 6.4.1-1

Find
To verify the Divergence theorem for the given vector field and the given domain as per Figure 6.4.1-1 We need to prove that : $$\int\limits_{\Omega }{\overline{\nabla }.}\overline{q}d\Omega =\oint\limits_{\Gamma }{\overline{q}.\overline{n}}d\Gamma $$ $$\displaystyle (Eq. 6.4.1-2) $$

Solution
We have, $${{q}_{x}}=-{{y}^{2}} $$  and secondly  $${{q}_{y}}=-2xy$$ as the given vector field over the domain given by Figure 6.4.1-1 Now we can write - $$\overline{\nabla }.\overline{q}=\frac{\partial {{q}_{x}}}{\partial x}+\frac{\partial {{q}_{y}}}{\partial y}=0-2x=-2x$$ Integrating over the given domain, we get LHS for $$ Eq.6.4.1-2 $$ as -

$$\int\limits_{\Omega }{\overline{\nabla }.}\overline{q}d\Omega = $$ $$\int\limits_{-1}^{1}{\int\limits_{-1}^{1}{2x.dydx=}}\int\limits_{-1}^{1}{\left( 2x.\left[ y \right]_{-1}^{1} \right)}dx=-2\left[ {{x}^{2}} \right]_{-1}^{1}=0 $$

Thus : $$\displaystyle (Eq. 6.4.1-3) $$ Also we have for RHS of$$ Eq.6.4.1-2 $$ as - $$\oint\limits_{\Gamma }{\overline{q}.\overline{n}}d\Gamma $$ = $$\int\limits_{AB}^ – {2xy.}d\Gamma +\int\limits_{BC}^ – {(-{{y}^{2}}).}d\Gamma +\int\limits_{CD}^ – {(-2xy).}d\Gamma +\int\limits_{DA}^ – {{{y}^{2}}.}d\Gamma =\int\limits_{-1}^{1}{2xy.}dx-\int\limits_{-1}^{1}{({{y}^{2}}).}dy+\int\limits_{1}^{-1}{2xy.}(-dx)+\int\limits_{1}^{-1}{{{y}^{2}}.}(-dy) $$

Hence, RHS = $$\oint\limits_{\Gamma }{\overline{q}.\overline{n}}d\Gamma $$ = $$\int\limits_{-1}^{1}{2xy.}dx-\int\limits_{-1}^{1}{({{y}^{2}}).}dy-\int\limits_{-1}^{1}{2xy.}dx+\int\limits_{-1}^{1}{{{y}^{2}}.}dy$$ = $$ \left( 2y\int\limits_{-1}^{1}{x.}dx-\int\limits_{-1}^{1}{({{y}^{2}}).}dy-2y\int\limits_{-1}^{1}{x.}dx+\int\limits_{-1}^{1}{{{y}^{2}}.}dy \right)=0$$

Thus :

$$\displaystyle (Eq. 6.4.1-4) $$ From $$Eq. 6.4.1-3 $$ and $$ Eq. 6.4.1.-4 $$, we can say that -

This proves that the given vector field as per $$Eq. 6.4.1-1 $$in the stated domain of Figure 6.4.1 satisfies the Divergence Theorem

=Problem 6.10 =

2D space with boundary convection
Consider a heat conduction problem in 2-dimensional space with boundary convection as shown in the below Figure 6.10.1.

Boundary Conditions
Above 2D space has temperature $$ T(x,y)$$ as space unknown to solve for. We can state it as a boundary value problem with following boundary conditions specified at 3 different kind of boundaries as : $${\Gamma _g}$$, $${\Gamma _h}$$ and $${\Gamma _H}$$

Essential BC on $${\Gamma _g}$$
$$\displaystyle (Eq. 6.10.1) $$

Note : $$T (x,y)$$ can also can be defined as a varying quantity along $${\Gamma _g}$$

Natural BC on $${\Gamma _h}$$
Refer to the Figure 6.10.3 below. We specify the heat flux on boundary $${\Gamma _h}$$ as $${q_n} = {q_0}$$.

This is as we have : $$ {q_n} = \overline q .\overline n $$ with $$\overline n $$ as the unit normal vector on $${\Gamma _h}$$ Hence we can also write : $${q_n} = ( - k.\overline \nabla T).\overline n  =  - k(\overline \nabla  T.\overline n )$$ OR we then have :

$$\displaystyle (Eq. 6.10.2) $$ Note- $$ \nabla T$$ is positive as explained in the - Assumption- subsection below

We are to consider only the normal component of $$\overline q $$, i.e. $$ {q_n}$$, for heat transactions as that is the only component causing heat flow in or out of the element. The tangential component of $$\overline q $$, i.e. $$ {q_t}$$, is just circulating over the boundary.

Assumption
$$\overline n  $$ is unit vector pointing in the outward direction. This implies that $$ \nabla T$$ is positive and temperature increases as we go out away from the surface towards inside of the body. Also we note that $$\overline n  $$ is a vector quantity normal to the $$ T(x,y)$$ contours in 2D space. As $$ T(x,y)$$ increases from inside to outside, heat flux $${q_n}$$ goes into the volume from outside of the boundary $${\Gamma _h}$$ This implies that when $${q_0}$$ is positive, heat comes into the system

Convection BC on $${\Gamma _H}$$
We have as the convection boundary where heat transfer occurs via heat convection Hence we can say that the normal component $${q_n}$$ of the heat flux $$\overline q $$ equals the heat transfer due to heat convection at $${\Gamma _H}$$. Hence we write-

$${q_n} = h(T - Ta)$$ where, we have - $$T$$ : Temperature at the surface of $${\Gamma _H}$$ $${T_a} = {T_\infty } =$$ Say, Ambient temperature $$h$$ :Coefficient of heat convection at $${\Gamma _H}$$ Thus :

$$\displaystyle (Eq. 6.10.3) $$

Deriving the Equilibrium Equation for 2D- heat transfer
We first derive the equilibrium equation using elemental plane surface as shown in Figure- 6.10.3 below. The elemental surface has length as 'dx' and height as 'dy'. The inflow and outflow of heat is shown symbolically with arrows at various sections.

Note- Heat flux ' $$ \underline q $$ ' is a vector quantity in real 2D space defined per unit surface area and also $$f = $$ Heat generation per unit volume for the given elemental surface We assume unit dimension along the direction perpendicular to the screen or paper

Now the equilibrium of heat can be written in following manner -

Now we use | Taylor Series Expansion technique and can then write -

$$ {({q_x})_{x + dx}} = {({q_x})_x} + {\left( {\frac} \right)_x}dx + higher\_order\_terms$$ OR we can say with the higher order terms neglected :

Similarly we can write for y- direction - And now $$ Eq. 6.10.4 $$ can be modified as below - Canceling out $$dxdy $$ throughout as we have used it for elemental surface and it has to be true over the entire 2D space, $$ Eq. 6.10.8 $$ can be written with different notation in the form of Divergence of $$\overline q $$ as - Note that $$ Eq. 6.10.9 $$ is valid for 2D as well as 3D heat transfer with $$\overline q $$ as 2D or 3D vector respectively in real 2D or 3D space. Here we are solving for temperature $$T(x,y)$$ so we use 2D space with $$T(x,y)$$

Constitutive Equation for 2D- heat transfer
Here we use the Fourier Law for 2D space which is stated as : $$\overline q  =  - \left[ k \right]\overline \nabla  T $$ It can be written as - Note- Usually, $$=0 $$ and $$=0 $$ for most of the materials as we do not see flow of heat perpendicular to the direction of the temperature gradient line. Also, we have : $$= $$ for isotropic materials and $$ \ne  $$ for anisotropic materials We consider isotropic materials here hence we can assume : $$= $$. Hence $$ Eq. 6.10.10 $$ now becomes - Hence we write then :

Governing Equation for 2D- heat transfer
Using $$ Eq.6.10.9 $$ and $$ Eq. 6.9.10 $$ we can say that :

Thus the strong form for 2D - Heat Transfer can be written as follows-

$$\displaystyle (Eq. 6.10.11) $$

Deriving the Weak Form for 2D- heat transfer with boundary convection
Let us select the weighting function $$ w(x,y) $$ such that $$ w(x,y)=0$$ at boundary $$ {{\Gamma }_{g}} $$ where the essential boundary condition is applied. This is actually obvious to say as the essential boundary condition is specified at $$ {{\Gamma }_{g}} $$, the value of $$ T (x,y)$$ will be constant there and will no longer require any weighting function multiplication to approximate it in the given domain. Also this will eliminate the unknown flux at the boundary $${{\Gamma }_{g}} $$

Integrating over volume (for general 3D - case) after multiplying with the weighting function w(x), in the form of weighted residual, we get - Note- For 2D- case : Put $$ dz $$ = $$1$$ which will give us the Area Integral Now we use Green's Theorem to perform the integration by parts. Our aim to get rid of the derivative on $$ \overline{\nabla }.(k\overline{\nabla }T)$$ and a derivative should appear on $$ w(x) $$ We state the general Green's Theorem for scalar fields $$ \varphi $$, T - is as follows -

Here, $${{\nabla }^{2}} $$ serves as Laplacian or Divergence of the gradient. Also, $$\overline{n} $$ is the unit normal at boundary and is variable at each point on the boundary. Thus ,we have a volume integral converted to a volume plus a surface integral Hence, $$ Eq. 6.10.12 $$ can now be modified as follows - This is general equation for 3D as well as 2D case and can be converted to specific case depending upon whether the gradient is 2D or 3D Now again we note that in the weak form the natural boundary conditions get pulled in automatically, for example- the surface integral in above equation will be split into integral over the 3 boundaries - $${{\Gamma }_{g}} $$, $${{\Gamma }_{H}} $$ , and$${{\Gamma }_{h}} $$ , as follows :

Note- In the above Equation, the first term on the RHS will vanish as the weighting function $$ w (x) $$ vanishes on $$ {{\Gamma }_{g}}$$ as explained. Also here we flipped the signs for convection boundary condition on boundary on $${{\Gamma }_{g}} $$ Now, we put $$ Eq. 6.10.15 $$ into $$ eq. 6.10.14$$ and keep the terms required to get the stiffness matrix $$ \left[ K \right]$$ along with the terms involving temperature on LHS and all the terms with known values on the RHS.

$$\displaystyle (Eq. 6.10.16) $$ $$ Eq. 6.10.16 $$ is the weak form for the given 2D Heat Transfer domain with convection boundary specified as $$ {{\Gamma }_{H}}$$ It is the continuous form of the weak equation and we need to convert it into a discrete form using the method of Desensitization in FEM. Also note that - 1. In $$ Eq. 6.10.16 $$, temperature is a continuous function of field parameters (x,y). 2. This equation is good for 2D as well as 3D domains. 3. We can handle anisotropic materials by converting conductivity 'k' into matrix form as explained.

Parametric space for 2D heat transfer
We select the following perfect square as the parametric element for this case. The coordinate system is s-t and nodes are denoted by numbers in anticlockwise manner

Shape Functions for 2D Heat transfer
We derive the shape functions for this 2D - Heat transfer case in following format -

System of equations in descritized form
Now the temperature distribution in 2D space can be written as follows with subscript 'e' denoting to an element i.e. temperature at a point (x,y) in the 2D space is a sum over that of the elemental nodes. Also here $$ N_i $$ denotes shape function and $$ T_i $$ denotes nodal temperature values.

This can be written in matrix form as follows where the nodal values matrix is shown by $$\left\{ {{X}_{e}} \right\} $$

Also applying the same for the weighting function, we can write : This can be written in matrix form as follows -

We note that we have got gradient terms in the weak form as seen in $$Eq. 6.10.16 $$. The $$\overline{\nabla }T $$ can be written as : Here $$ \left[ B \right]$$ denotes the matrix of derivatives of the shape functions as shown in below $$ Eq. 6.10.22 $$.

Similar equation can be written for the weighting function as follows -

Now the LHS of the weak form in $$ Eq. 6.10.16 $$ can be written as for a single element -

Hence, LHS becomes for $$ Eq. 6.10.16$$ as :

Depending upon various boundary conditions the RHS of $$ Eq. 6.10.16 $$ can be written as $$\left[ {{F}_{e}} \right]{{\left\{ \partial {{X}_{e}} \right\}}^{T}} $$ Hence the discrete form Equation for 2D Heat Transfer case becomes -

where, $$ \left[ {{K}_{e}} \right]=\int\limits_{V}{{{\left[ B \right]}^{T}}\left[ K \right]}\left[ B \right]dv$$

=Contributing Members=

title: User talk:Anand ST14