材料力學-第二章

1、一、工程實例一、工程實例 三、變形特點三、變形特點二、外力特點二、外力特點四、外力計算簡圖四、外力計算簡圖FFFFmmFF一、求內(nèi)力一、求內(nèi)力 mmFFNmmFFFNmmFFmmFFNmFmFNmFFmmFFNmFm二、軸力圖二、軸力圖x xF FN NO OC CA AB BD D600600300300500500400400E E40kN40kN55kN55kN25kN25kN20kN20kNC CA AB BD D600600300300500500400400E E40kN40kN55kN55kN25kN25kN20kN20kNC CA AB BD DE E40kN40kN55kN5

2、5kN25kN25kN20kN20kNF FR RA AkN100202555400RR AAxFFFC CA AB BD DE E40kN40kN55kN55kN25kN25kN20kN20kN1 10R1N AFF)()kN(10R1N AFF F F40kN40kNF FN2N220kN20kNC CA AB BD DE E40kN40kN55kN55kN25kN25kNF F2 2040R2N AFF)()kN(5040R2N AFFF FN3N320kN20kN25kN25kNC CA AB BD DE E40kN40kN55kN55kN25kN25kN20kN20kNF F3 30

3、 02 20 02 25 53 3 NF)()kN(53N F20kN20kNF FN4N440kN40kN55kN55kN25kN25kN20kN20kNF F4 4)(kN)204N F505010105 52020+ + +C CA AB BD D600600300300500500400400E E40kN40kN55kN55kN25kN25kN20kN20kN(kN)50Nmax F 一、橫截面上的正應力一、橫截面上的正應力FFabcdF FF Fa ab bc cd da b c d F F F FN NAFN F Fk kk k F F coscos AFAFp二、二、 斜截面上

4、的應力斜截面上的應力F Fk kk kF Fp p AFp cosAA FF 2coscosp sinsin22p F Fk kk k F FF Fk kk kx xn n p p max2 2 max2 2 min0 00 0 ,2coscospsinsin22p x xn nF Fk kk k F FA AB BC CF FF F30003000400040003703702402402 21 1kNN50501 1 FFkNN1 15 50 03 32 2 FFF FA AB BC CF FF F30003000400040003703702402402 21 150kN50kN150k

5、N150kNMPa.N/m.2N8 87 70 01 10 08 87 70 02 24 40 02 24 40 05 50 00 00 00 06 61 11 11 1 AF MPa.N/m.2N1 11 110101 11 137370 037370 01500001500006 62 22 22 2 AF max 一、實驗方法一、實驗方法二、拉伸試驗二、拉伸試驗 d dl l標距標距F FO Ol le ef fh ha ab bc cd dd dg gf fl l0 0 p p E p f fO Of fh ha ae p p f fO Of fh ha ab b e es s s b

6、 b b e e p p f fO Of fh ha ab bc ce e s s b b e e p p f fO Of fh ha ab bc ce e %1 10 00 01 1 lll %1001001 1 AAA a ab bc ce ef fO Og gf fh hd dd dpe a ab bc cd de ef fO Od dg gf fh hd ds s 0.20.2e e 0.20.2% % tan E2 20 0. oO O / /MPaMPa/%/%e e dh0 03 35 51 1. dh s sO O e e O O /%/%e e nu ssn bbn Nmax

7、max AFNmax FA NmaxAF AFmaxN ABCF1mABCF1mFAxyF FA Ax xy y0 030300 01 1 FFFysinN0 00 01 12 2 cos30NNFFFxFFFF7327321 12 22 21 1.NN 22mm6 62 26 61 11 10 02 28 86 60 02 21 14 43 30 01 10 02 21 17 72 22 21 10 08 86 6 AAAFmaxN FFFF732. 122N1N kN.N2 24 43 36 69 91 11 1 AF kN.N2 20 04 48 86 62 22 2 AF kN.N6

8、61 18 84 42 21 11 1 FFkN.N7 72 28 80 07 73 32 21 12 22 2 FF2 2a aa aF FA AB BD DC C2aaFABDCCDCDF FA AC CB BFFMCDA230N /N232119MPa4CDFFAd N AFCDCD F F=33.5kN=33.5kN2 2a aa aF FA AB BD DC CCDCDF FA AC CB B2 23 3FAFCD N N AFCDCD/N 2 23 3FFACD / 2324dF FFbh 一、縱向變形一、縱向變形h1 lllll 1 1 二、二、橫向變形橫向變形三、泊松比三、泊松

9、比bbbbb 1 1 FFbhh1 bbb 1四、胡克定律四、胡克定律 AFN ll E EAlFlN F F1 1F F2 2F F3 3l l1 1l l2 2l l3 3A AB BC CD DF F1 1F F2 2F F3 3l l1 1l l2 2l l3 3A AB BC CD DF F1 1F FN1N1)(kN2001N1N1 FFFF2F1FN2F1F2F3l1l2l3ABCD)(kN1502N2N21 FFFFFFN3)(kN5003NR3N FFFDFN2 =-15kN （-）FN1 =20kN （+）FN3 =- 50kN （-）15+-2050F1F2F3l1l2l

10、3ABCDF FN2 N2 =-15kN ( - )=-15kN ( - )F FN1 N1 =20kN =20kN （+ +）F FN3 N3 =- 50kN ( - )=- 50kN ( - )F1F2F3l1l2l3ABCD)(MPa.N 8 81761761 11 1AFAB )(MPa.N 6 674742 22 2AFBC )(MPa.N 5 51 11 10 03 33 3AFDC F1F2F3l1l2l3ABCDm102.534-N 1 11 11 1EAlFlABm101.424-N 2 22 22 2EAlFlBCm101.584-N 3 33 33 3EAlFlCD-0.

11、3mm BCCDBllumm10-0.474- CDBCABADllllABC12 ABC12 , ,FyA12 x0coscos00sinsin02N1N1N2N FFFFFFFyx cos22N1NFFF AABC12 ABC12 cosNEAFlEAlFll2 21 11 12 21 1 ABC12 A A2 2A1Al 1 12AAAAAA )(mm293. 1cos2cos21 EAFllAAAF FA A1 12 2x x3030y yA A1 1l 12m2mA AB BC CF F30301 12 2FFFF732. 122N1N A A2 2l 2mm.mm.NN765765

12、0 01981981 12 22 22 22 22 21 11 11 11 11 1 EAlFlAAEAlFlAA3030AA1A2l 1l 2A30A1 1l 12mABCF301 12 2A2 2l 2A330cos1222llAAAAAA 30sin30tantan3012232llAAAA mm.)()(7 78 83 32 23 32 22 22 23 3 AAAAAA ()dWFdl10()lWFdl在在 范圍內(nèi)范圍內(nèi),有有p12WF l應變能（應變能（ ）：固體在外力作用下，因變形而儲）：固體在外力作用下，因變形而儲 存的能量稱為應變能。存的能量稱為應變能。V12VWF l212

13、2FlF lFEAEAFlll()dlFl1FFdFO1l 一、靜定與超靜定問題一、靜定與超靜定問題二、超靜定問題求解方法二、超靜定問題求解方法 C CA AB BD DF F 1 12 23 3三、一般超靜定問題舉例三、一般超靜定問題舉例x xy yF FA A 2 21 10 0NNFFFx 0coscos03N2N1N FFFFFy C CA AB BD DF F 1 12 23 3x xy yF FA A C CA AB BD D 1 12 23 33 3l A A1 12 23 3 C CA AB BD DF F 1 12 23 3C CA AB BD D 1 12 23 31 1l

14、 cos3 31 1ll 1 11 11 11 1EAlFlN 3 33 33 33 3AElFl cosN 2 23 33 33 31 1cosNNAEEAFF C CA AB BD DF F 1 12 23 33 3l A A1 12 23 3 1 1l 2 23 33 32 21 12 21 1coscosNNAEAEFFF 2 21 1NNFF 0 03 32 21 1 FFFFNNNcoscos 2 23 33 33 31 1cosNNAEEAFF 2 23 33 33 32 21 1cosNAEEAFF A AB BC CF F3 3a aa al l2 21 1A AB BC C

15、F F3 3a aa al l2 21 1F FA AB BC C3 3a aa a2 21 1F FN1N1F FN2N2F FN3N3F F 0 xF0 xF 0 0yF03N2N1N FFFF 0 0BM0 02 22 21 1 aFaFNNA AB BC CF F3 3a aa al l2 21 1ABCl 3l 2l 1A AB BC C3 32 21 12 23 31 12 2 lll 1 11 11 11 1EAlFlN EAlFl3 33 3N EAlFl2 22 2N 2 23 31 12 2NNNFFF ABCF3aal21ABCl 3l 2l 1ABC321N2N3N1F

16、FF2 2 0 0 xF0 0 FFFFN3N2N102N2N1 aFaF6/53/6/N3N2N1FFFFFF 一、溫度應力一、溫度應力A AB B A ABBA AB BBB0 0 lA AB BA AB BA ABB0 0 FTlllEAlFlBFR lTllT EAlFlTBlR TEAFlBR TEAFlBTR BBA AB B A二、裝配應力二、裝配應力A AB BC CD D 2 21 13 3l l AA AB BC CD D 2 21 13 3l ll 3l 13 3l1l3l 1cosl cos1 13 3ll3 33 33 31 11 11 1AElFlAElFlN3N1

17、cos AA AB BC CD D 2 21 13 3l ll 3 l 1 2 21 11 13 33 3cosN1N3AElFAElF0coscos0sinsinN2N1N3N2N1 FFFFF A AB BC C1 12 2a aa aB B1 1A A1 1C C1 1l l3 3C C1 1CCl l3 3C C1 1CCABC12B1C1A1ell 3 31 1a aa ax xEAlFl1 11 1N1 3 33 33 3AElFlN3 AElFeAElFN1N3 3 33 3CCAABB0 0 N2N1N3FFFN2N1FF max max max Kmax max 一、基本概念

18、和實例一、基本概念和實例 （F FF FF FF FF FF F m mF FF FA AB Bd dd dd d1 1d d1 1n nn nF FF F n nn nF FF Fm mm mF F剪切面剪切面F FS S二、剪切的應力分析二、剪切的應力分析 F FF Fm mm m00S FFFxFF SAFS m mm mF F剪切面剪切面F FF Fm mm m AFSnu u 三、擠壓的應力分析三、擠壓的應力分析FFFFF FF F bsbsbs AFAFbsbs h hbsAdh四、強度條件的應用四、強度條件的應用 bsbs FA bsbs SFA AFbsbs SFA u b bh hl lM Me eF FM Me ee2MdF kNe5 57 71 10 07 70 01 10 02 22 22 23 33 3 dMF MPa6 .281010020105763SblFAFFF S bsbsbsMPa. 3 39 95 51 10 06 61 10 00 01 10 05 57 72 26 63 3hlFAFM Me eF Fb bh hl ld d1 1F FF FA Ad dd d1 1B Bd dd d1 1F FF FA Ad dd d1 1B Bd dd d2F2FF F剪切面剪切面d d