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当前位置:首页 > 建筑/环境 > 综合/其它 > 2章习题参考答案材料力学课后习题题解
2.1试求图示杆件各段的轴力,并画轴力图。-+++FFFF20kN30kN50kN40kN40kN10kN20kN(2)(1)FN图图NF15kN15kN20kN10kN(4)10kN5kN10kN30kN+---FN图l(5)qFFFql2.2已知题2.1图中各杆的直径d=20mm,F=20kN,q=10kN/m,l=2m,求各杆的最大正应力,并用图形表示正应力沿轴线的变化情况。答(1)63.55MPa,(2)127.32MPa,(3)63.55MPa,(4)-95.5MPa,(5)127.32MPa15kN15kN20kN10kN(4)31.85MPa15.82MPa+---Fs图31.85MPa95.5MPal(5)qF+127.32MPa63.69MPa2.4一正方形截面的阶梯柱受力如题2.4图所示。已知:a=200mm,b=100mm,F=100kN,不计柱的自重,试计算该柱横截面上的最大正应力。4m4mabF题2.4图FFFFFFFFFFF4m4m解:1-1截面和2-2截面的内力为:FN1=-F;FN2=-3F相应截面的应力为:3N11213N22221001010MPa100300107.5MPa200FAFA最大应力为:max10MPa2.6钢杆受轴向外力如图所示,横截面面积为500mm2,试求ab斜截面上的应力。3020kNobaababpαsαατFNFN=20kNoNN0cos30==FFpAAααo2oN03cos30cos302010330MPa5004FpAαα解:3oooN020103sin30cos30sin3017.32MPa5004FpAαατ-+20kN20kN20kNⅠⅡⅢ20kN20kN1m1m2m12320N0N20NNNNFkFkFk2.8图示钢杆的横截面积A=1000mm2,材料的弹性模量E=200GPa,试求:(1)各段的轴向变形;(2)各段的轴向线应变;(3)杆的总伸长。解:轴力图如图所示41119624333962011020010100010020221020010100010NNFlLmEALmFlLmEAIIIIII0.1mm00.2mm0.1mmllll41243100210LmLmLm4411122244333101010210102LmlmLlLmlmFF(a)ABCDCAFFFFFFFFABCBACADBDACADDFCBFBDFABF2.10图示结构中,五根杆的抗拉刚度均为EA,杆AB长为l,ABCD是正方形。在小变形条件下,试求两种加载情况下,AB杆的伸长。解(a)受力分析如图,由C点平衡可知:F’AC=F’CB=0;由D点平衡可知:F’AD=F’BD=0;再由A点的平衡:xAB=0:=FFFABABFlFlLEAEA因此CFFCBACFFAFFFABACADFFADBDDFCBDF(b)AFCBFFABBD(b)受力分析如图,由C点平衡可知:0:0:22cos45,2xACBCyoACACFFFFFFFF由D点平衡可知:0:0:22xADBDyADFFFFFFCFFCBACFFAFFFABACADFFADBDDFCBDF(b)AFCBFFABBDABABFlFlLEAEA因此再由A点的平衡:0:cos450;oxACADABABFFFFFF2222ACADFFFFCFFCBACFFAFFFABACADFFADBDDFCBDF(b)AFCBFFABBDa2m1.5m1mBA②①F2.12图示结构中,水平刚杆AB不变形,杆①为钢杆,直径d1=20mm,弹性模量E1=200GPa;杆②为铜杆,直径d2=25mm,弹性模量E2=100GPa。设在外力F=30kN作用下,AB杆保持水平。(1)试求F力作用点到A端的距离a;(2)如果使刚杆保持水平且竖向位移不超过2mm,则最大的F应等于多少?解:受力分析如图a2mBAFFFN1N2220:2012NANMFFaFFa1120:220,2NNBaMFaFFFN11N22121122121122FlFlLLEAEAF2-alFal2EA2EA()121122=F2-alFalEAEAd1=20mm,E1=200GPa;d2=25mm,E2=100GPa。9269262241.542001020101001025101.5,1.07911.08m2025--2-aa2-a2aaππ12N222222229262222m2m410010251044181.95kN1.081maxmaxLLFlFalLEA2EAEAFalπ1m45o30oABCFFAo30o45FFABAC2.15图示结构中,AB杆和AC杆均为圆截面钢杆,材料相同。已知结点A无水平位移,试求两杆直径之比。解:0:cos45cos3002332xooABACABACABACFFFFFFF1m45o30oABC,,,,,AA45o30oA,cos45cos30cos303cos452ooABACoABACACoLLLLL由两杆变形的几何关系可得2;222yAByACABACLLLLLL'2sin4521sin302oAByoACyLAALAALAALAA222222222233321.06224221.03ACACABABABACACACABABABACABABABACACACABACFLFLAAFLFLdddFLdFLdd22ABACLL32ABACFF32ABACLL2.20图示结构中,杆①和杆②均为圆截面钢杆,直径分别为d1=16mm,d2=20mm,已知F=40kN,刚材的许用应力[σ]=160MPa,试分别校核二杆的强度。解:受力分析如图21o30o45FFo30o45FF1212120:sin45sin300(1)0:cos45cos300(2)xooyooFFFFFFF(1)+(2)可解得:F2=29.3kN;F1=20.7kN杆①和杆②都满足强度要求。d1=16mm,d2=20mm,[σ]=160MPaF2=29.3kN;F1=20.7kN21122112222222420.7420.710103MPa[]160MPa3.1416429.3429.31093.3MPa[]160MPa3.1420FAdFAd2.24图示结构,BC杆为5号槽钢,其许用应力[σ]1=160MPa;AB杆为100×50mm2的矩形截面木杆,许用应力[σ]2=8MPa。试求:(1)当F=50kN时,校核该结构的强度;(2)许用荷载[F]。解:受力分析如图ACB60oo60BFFFBABCF0:sin60sin300(1)0:cos30cos600(2)yooBCBAxooBABCFFFFFFF联立(1)和(2)解得:FBC=25kN;FBA=43.3kN。查型钢表可得:ABC=6.928cm2,FBC=25kN;FBA=43.3kN;ABC=6.928cm2,[σ]1=160MPa;AAB=100×50mm2;[σ]2=8MPa。311222251036.1MPa[]160MPa6.9281043.38.66MPa[]8MPa10050BCBCBABAFAFA杆BC满足强度要求,但杆BA不满足强度要求。22[][];[][]81005040NBABABABAFFAkA将[FBA]带入(1)、(2)式中求得许用荷载[F]=46.2kNFABDC1m1m0.75m2.25图示结构中,横杆AB为刚性杆,斜杆CD为直径d=20mm的圆杆,材料的许用应力[σ]=160MPa,试求许用荷载[F]。解:CD=1.25m,sinθ=0.75/1.25=0.6FABD1m1mFFFDCAxAy¦ΘADCDCM0:2sin102100.63=-?q?==åFFFFF32262634104010[]16033201016032010[]15.14010DCDCFFFAdFkN32262634104010[]16033201016032010[]15.14010DCDCFFFAdFkNFABDC1m1m0.75m2.25图示结构中,横杆AB为刚性杆,斜杆CD为直径d=20mm的圆杆,材料的许用应力[σ]=160MPa,试求许用荷载[F]。解:CD=1.25m,sinθ=0.75/1.25=0.6FAB1m1mFFFDCAxAy¦ΘDADCDC0:2sin102100.63=-?q?==åMFFFFF232626341034010[]1603201016032010[]15.14010DCDCFFAdFFkNDC103=FFd=20mm[σ]=160MPa钢木FBACAFFFACABa2.27图示杆系中,木杆的长度a不变,其强度也足够高,但钢杆与木杆的夹角α可以改变(悬挂点C点的位置可上、下调整)。若欲使钢杆AC的用料最少,夹角α应多大?解:答45o0:sin0yACFFFACACACACAC[][]sin/cosFFAlaACACACACACAC[][]sincos[]sin2FFa2FaV=Al杆AC的体积:钢杆AC的用料最少,则体积最小,有:ACACAC[]sin/cosFAlaosin21;452.37图示销钉连接中,F=100kN,销钉材料许用剪切应力[τj]=60MPa,试确定销钉的直径d。解:dFFFFF22350N24[]4501032.6mm3.1460ssjFFkFd2.39图示的铆接接头受轴向力F作用,已知:F=80kN,b=80mm,δ=10mm,d=16mm,铆钉和板的材料相同,其许用正应力[σ]=160MPa,,许用剪切应力[τj]=120MPa,许用挤压应力[σbs]=320MPa。试校核其强度。解:20N4sFFkbdFFFFF/4F/4F/4F/4F/4F/4FF/4F/43Fs123123/4==31.25MPa[]-)3/4==125MPa[]-2)==125MPa[]-)FbdFbdFbd(((bdFFFFF/4F/4F/4F/4F/4F/4FF/4F/43Fs123[σ]=160MPa20N4sFFkb=80mm,δ=10mm,d=16mm;[τj]=120MPa,[σbs]=320MPa323bs4201099.5[]3.14162010===125MPa[]1610sjjsbsFMPaAFd
本文标题:2章习题参考答案材料力学课后习题题解
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