彈性力學(xué)課件六

第三章平面問(wèn)題的直角坐標(biāo)解答3—1逆解法與半逆解法3—2矩形梁的純彎曲3—3位移分量的求出3—4簡(jiǎn)支梁受均布荷載3—5楔形體受重力和液體壓力SolutionofPlaneProblemsinRectangularCoordinatesInverseMethodandSemi-InverseMethodPureBendingofaRectangularBeamDeterminationofDisplacmentsBendingofaSimpleBeamUnderUniformLoadsTriangularGravityWall3—1逆解法與半逆解法多項(xiàng)式解答按應(yīng)力求解平面問(wèn)題時(shí)，如果體力是常量，則歸結(jié)為由求解應(yīng)力函數(shù)，再求應(yīng)力分量。關(guān)鍵是求應(yīng)力函數(shù)（x，y）解的方法:逆解法、半逆解法一、逆解法與半逆解法InverseMethodandSemi-InverseMethodSolutionbyPolynomialsInverseMethodandSemi-inverseMethod1、逆解法：先設(shè)定各種形式、滿足相容方程的應(yīng)力函數(shù)，然后求應(yīng)力分量，再根據(jù)邊界條件來(lái)考察這些應(yīng)力分量對(duì)應(yīng)什么樣的面力，從而得知所設(shè)定的應(yīng)力函數(shù)可以解決什么樣的問(wèn)題2、半逆解法：針對(duì)所要求解的問(wèn)題，根據(jù)彈性體的邊界條件，先設(shè)定一部分或全部應(yīng)力分量函數(shù)，從而導(dǎo)出應(yīng)力函數(shù)，再考察這個(gè)應(yīng)力函數(shù)是否滿足相容方程，應(yīng)力分量是否滿足邊界條件，若滿足，這個(gè)結(jié)果就是正確的，若不滿足，還須另作假設(shè)，重新考察二、多項(xiàng)式解答（逆解法）1、一次式?。▁，y）=a+bx+cy無(wú)論a、b、c取何值，均能滿足相容方程應(yīng)力分量：Inthissection,weshallusetheinversemethodtoobtainthesolutionofsomesimpleplaneproblemsbystressfunctionsintheformofpolynomials,assumingthattherearenobodyforcesacting,i.e.,X=Y=0Whenthestressfunctionistakenasapolynomialinlinearform,（x，y）=a+bx+cy,thecompatibilityequationissatisfiedforallvaluesoftheconstantsa,bandc.Thestresscomponentsare根據(jù)邊界條件考察應(yīng)力分量所對(duì)應(yīng)的面力結(jié)論（1）線性應(yīng)力函數(shù)對(duì)應(yīng)無(wú)體力、無(wú)面力、無(wú)應(yīng)力的情況（2）在平面問(wèn)題的應(yīng)力函數(shù)上加上一個(gè)線性函數(shù)，并不影響應(yīng)力面力為零無(wú)論任何邊界Thestressboundaryconditionsalwaysgiveregardlessoftheshapeofthebodyandthecoordinateaxeschosen.Thus,weconcludethatAlinearstressfunctioncorrespondstothecaseofnosurfaceforcesandnostressThesuperpositionofalinearfunctiontothestressfunctionforanyproblemdoesnotaffectthestresses.Whenthestressfunctionistakenasapolynomialoftheseconddegree,（x，y）=ax2+bxy+cy2Thecompatibilityequationisalsosatisfiedforallvaluesoftheconstantsa,bandc.Thestresscomponentsare：Inthecaseof

=ax22、二次式（polynomialoftheseconddegree）

根據(jù)邊界條件考察應(yīng)力分量所對(duì)應(yīng)的面力xy板左右兩側(cè)無(wú)面力上下兩側(cè)面力為2aWhensuchstressesexistinarectangularplatewithitsedgesparalleltothecoordinateaxes左右兩側(cè)：Leftandright：上下兩側(cè)：Upsideandunderside考察岸：

=b飄xy欄（In葉t蓬he閑c烈as蜂e珍of=曬bx帝y）結(jié)論根據(jù)斯邊界健條件罪考察躲應(yīng)力翠分量厲所對(duì)潤(rùn)應(yīng)的邀面力

=ax2能解決矩形板在y方向受均布拉力（a0）或者受均布?jí)毫Φ膯?wèn)題（a0）Thestressfunction=ax2cansolvetheproblemofuniformtension(a>0)oruniformcompression(a<0)ofarectangularplateinydirection.應(yīng)力分量：thestresscomponentsareWh火en般s營(yíng)uc膊h榮st廢re皮ss未e泊xi努st攜i急n迷a臨re魄ct隆an浙gu傻la飛r時(shí)pl誓at莫e,貓t窩he并re欠w資il厚l蝕be向u隆ni碗fo伸rm蚊ly遙d拼is稼tr倉(cāng)ib絹ut清e(cuò)d處s妖he事ar新in傭g醒fo重rc填es南o貨n腥al權(quán)l(xiāng)鹿th金e佩fo冒ur痰e緣瑞dg深es爐.xy左右兩側(cè)：板右側(cè)受剪力-b作用；板左側(cè)受剪力+b作用上下兩側(cè)：板下側(cè)受剪力-b作用；板上側(cè)受剪力+b作用Le梯ft摸a當(dāng)nd鋼r納ig矩ht：Up態(tài)si睛de矛a仁nd腹u繳nd辯er多si舉de：結(jié)論考察順最后汁一項(xiàng)墾：

=c至y2

=bxy能解決矩形板受均布剪力的問(wèn)題Hence,thestressfunction=bxycansolvetheproblemofarectangularplateinpureshear.

It霞i戰(zhàn)s據(jù)ea貍sy據(jù)t鋸o疫se艘e善th蕉e嶄st菜re馬ss跟f舒un漲ct染io界n=田cy2ca甩nso悉lv以e遼th架e煉pr干ob拐l(shuí)e覺(jué)m播of婆u卷ni俘fo騎rm釋t晚en劑si政on恒(職c>服0)混a欲nd絡(luò)u茫ni喘fo沾rm狗c僵om鍛pr唇es彼si籃on塘(劇c<削0)席o阿f筐a混re鎖ct舅an元gu螞la藥r菊pl桑at蔥e幣inxdi智re瓶ct技io呆n.應(yīng)力分量：Thestresscomponentsare根據(jù)錘邊界豪條件帶考察盞應(yīng)力眾分量堆所對(duì)糊應(yīng)的表面力xy左右兩側(cè)：上下兩側(cè)：板左右兩側(cè)面力為2c上下兩側(cè)無(wú)面力Le機(jī)ft漆a貓nd稻r擱ig科ht：Up艦si創(chuàng)de壘a嗽nd向u興nd懸er匠si化de：（x，斧y）涼=a投x2+b賓xy倘+c層y2能解布決什振么問(wèn)崗題？xy能解剖決矩曬形板物既受尚拉又概受剪起的情喊況Th池us簡(jiǎn),她th捐e聾pr多ob線le亂m庫(kù)wh嗚ic擇h羨th鑄e彩st穗re下ss代f飲un茶ct伯io瓶n(江x,郵y)炕=a印x2+b濕xy想+c麻y2ca套n罵so勺lv私e矛is?。▁，播y）律=a甲y3無(wú)論a取何值，均能滿足相容方程3、攤?cè)位晔絎h疼en材t崖he殖s奇tr灰es渠s準(zhǔn)fu效nc洪ti戰(zhàn)on神i圾s猜ta飲ke房誠(chéng)n桌as(x,輔y)班=a蕉y3,(1握)th剃e墨co干mp位at它ib收il辰it拆y比eq皆u(píng)a壇ti榴on亡i饑s糕al津so親s線at絲式is賠fi床ed肚f豎or詠a債ll陪v咸al擺ue敏s介of壞t己he第c脂on召st賴an至ta.應(yīng)力分量：Thestresscomponentsare(2材)xy根據(jù)佳邊界鴨條件待考察繳應(yīng)力唐分量旱所對(duì)旁應(yīng)的倆面力左右兩側(cè)：板左蓋右兩煌側(cè)受勻線性極分布惰的沿y方向擾的面轟力Fo絮r源a棗re殊ct佛an屯gu找la貞r掀pl者at錄e,秩w咐it添h釋co推or密di嬌na翻te悶a個(gè)xe蜘s享sh足ow顏n秒in筋F礙ig議.,Le限ft禾a返nd燈r娃ig折ht塘s平id口e上下兩側(cè)：上下弊兩側(cè)榆無(wú)面額力矩形松板左類右兩劃邊的史力合臣成為宴一個(gè)繡力偶消，所田以，

=a棋y3能解園決矩拒形板傻的純?nèi)澢葐?wèn)題筒。xyUp柜si雹de半a捧nd跟u逝nd弦er恭si歲deHe渣nc暴e,飼t拘he蒼s效tr鬧es縫s雨fu惠nc烘ti稅on(x,杜y)腥=a礦y3ca逗n阻so撈lv算e叮th勇e眨pr筋ob么le跳m哲of酷p裹ur詳e很be河nd他in增g預(yù)of雅a營(yíng)r黃ec覆ta泉ng激ul丸ar樂(lè)b伴ea棄m.Th難e男co恐ns溜ta攪ntaca兵n省be戀d病et理er膚mi次ne幻玉d獄by眨t漁he堅(jiān)b薯en末di暗ng灶m困om司en竿t.xyh/2h/2MMh1Forconvenience,weconsideronlyaunitwidthofthebeam.LetthebendingmomentontheunitwidthbeM,whichhasadimensionof[force][length]/[length],orsimply[force].Su龍bs豬ti左tu叉ti爽on巨o帳f吳th械e竭ex菌pr士es汗si擦on膊f承or

xin泄E退qs茫.(雀2)隨i吧nt稠o麻Eq渴s.書(3鬧)患yi稀el蜘dsOneitheroftheverticaledges,thesurfaceforcesmusthavearesultantforceequaltozeroandaresultantmomentequaltoM.Thisrequires(3)Th礙e滋fi紋rs會(huì)t叉eq秒ua查ti兆o(hù)n適i葡s家al最wa贊ys巾s膊at暫is螺fi圓ed婆w死hi致le秧t眾he睜s盼ec拾on夠d案eq遞ua臥ti躺on芽r序eq梯ui伸re待s6a=1馳2M/h3.Since：ThemomentofinertiaofthebeamsectionThus,Eqs.(2)become：(4)Eqs.(4)mayberewrittenas(5獸)Eq惕s(句5)況c乒oi瞇nc散id祥e灘co紛mp匙le堪te叛ly降w困it變h免th班e態(tài)el別em煌en憂ta暫ry虧s特ol甚ut蒼io擊n盛gi諷ve跨n黑in誼m撕ec誕ha仔ni齒cs搬o趙f托ma話te駕ri體al士s.No研te：th罰e蹲so風(fēng)lu創(chuàng)ti承on日(披5)校i瘡s花ex駕ac及t鐘on清ly冬w冤he支n惑th芽e兔su惡rf過(guò)ac久e骨fo塊rc呈es寄o騰n班th哪e脾en裂ds甩o呆f碰th鐘e懷be稿am道c乒on泊si引st池o麗f絲式no勇rm熊al軌t狗ra悟ct控io再ns肝p漫ro速po明rt擦io耗na趣l關(guān)toy.If各t施he抵c奸ou原pl務(wù)es牢o爛n麻th續(xù)e批en仆ds蜘a撇re臣a幟pp蛾li發(fā)ed滿i誼n健an都y碎ot巾he趟r恐ma健nn狂er勁,秀th窮is紛s拒ol咬ut功io州n晴is耕n爆o徹lo集ng亞er抹e扁xa糕ct潮,絞be罪ca剩us前e因it咱w標(biāo)il勾l億no眠t浴pr規(guī)ec別is己el界y王sa聲ti思sf削y溪th痛e括bo謠un凳da撈ry規(guī)c刑on漿di魂ti倆on但s勢(shì)on廳t膨he弓e粒nd咸s.Ac丸co距rd山in崗g抓to煤S篇ai乓nt晴-V策en匠an醋t’焰s境pr制in獸ci勿pl尖e,鞋h革ow抖ev刊er筒,仍it麥d對(duì)oe屬s唉re盟pr元e