《平面任意力系》PPT课件

1、FdFMMBB)()(10111FMMFF)(20222FMMFF)(0nnnnFMMFF)(iOiOFMMMOMRiiFFFRFRiFF )(iOOFMMxixixRxFFFFyiyiyRyFFFF22)()(iyixRFFFRixRFFiF),cos(RiyRFFjF),cos()(iOOFMM()()(32)OoiiiyiixMMFx Fy F) 13(),cos(),cos()()(22RyRRxRyxRFFjFFFiFFFF=0RF0OM0OM0RFROFM0OM0OMROFMddFMRoRRRFFF )33()()(iOORoFMMFM00oRMF)()()(22iOOyxRFMM

2、FFF)43(000oyxMFF000AyxMFF000BAxMMFBA,000CBAMMMCBA, 0 xF0000 0 xF0coscoscos321FFF 0yF0sinsinsin321FFF00AyMF00BAMMBA,mn33nm32 nm32 nm32 nm32 nm 在长方形平板的在长方形平板的O，A，B，C点上分别作用着有四个力：点上分别作用着有四个力：F1=1 kN，F2=2 kN，F3=F4=3 kN如图，试求以上四个如图，试求以上四个力构成的力系对力构成的力系对O点的简化结果，以及该力系的最后合成结点的简化结果，以及该力系的最后合成结果。果。求向求向O O点简化结果点简

3、化结果xxFFR30 cos60 cos432FFFkN 598. 0yyFRF30 sin60 sin421FFFkN 768. 0kN 794. 0 2R2RRyxFFFRF FOOMMmkN 5 . 030 sin3260 cos2432FFFm 51. 0RFMdO614. 0cosRRRFFxi , F789. 0 , cosRRRFFyjF1 .52Ri , F9 .37Rj , FRRFFRFmmN 93.78 cos)(nnnrFhFFMO cos)()()(nrnrFFMFMFMFMOOOOrqlxqFl021dqlxq qlxxqFh0dlh32q9m3m1.5m3.9m5

4、.7m3m907 .16 arctanCBABACBkN 1 .670 sinkN 9 .232 cos221R21RFGGFFFFFFyyxxxRFRFyRF3m9m1.5m3.9m5.7m3m90kN 4 .709)()(22RRyxFFFF945.0,cos328.0,cosRRRRFFFFyxjFiF16.19180,84.70,RRjFiFxRFRFyRF84.70yOxOOOMMMMRRRFFF84.7084.700RxOMFxFMMyyOORRFm514.3RyOFMxmkN 355 2m 9 .3m 5 .1m 3 211GGFMMOOFxRFRFyRF84.70RROORyx

5、yxMMxFyFxFyFF84.70yxkN 9 .232kN 1 .670mkN 355 20mkN 355 2kN 9 .232kN 1 .670yx45 0245 cos, 0045 sin, 0045 cos, 0lFlFMFFFFFFFCACAyyCAxxFkN 36.2222RAyAxAFFFkN 1045 sinkN 20245 coskN 28.2845 cos2FFFFFFFFFCAyCAxC45, 0 xF0 cosBAxFF, 0yF0 sin21BAyFFGFF , 0FMA0 sin cos221lFcFblFlGaFBB60kN 75. 0AxFkN 56. 3By

6、FkN 261. 0AyF60, 0 xF060 cos2FFAx0)(FMA060 sin)(212112llFlFMlFBy, 0yF060 sin21FFFFByAy60q45045 cos, 0FFFAxx045 sin, 0FqlFFAyy 045 cos2, 0MlFlqlMMAAF 707. 045 cosFFFAx 707.0FqlFAy 707.0212MFlqlMAq45q45FDFD, 0 xF0AxF, 0yF0DAyFFF , 0FAM0m 22MFABFD3060030 cos260 sin , 0030 sin60 cos , 0FqlFFFFFFFBAyyBAx

7、x3060 0430 cos360 sin22 , 0lFlFlqlMMFMBAA 0230 cos260 sin, 0lFlqllFFMBCmkN 37.10kN, 32. 2kN 89.32kN, 77.45AAyAxBMFFF306030600FGFAy0ABFACGMA, 0 xF0AxF, 0yF , 0FMA 一种车载式起重机，车重G1= 26 kN，起重机伸臂重G2 = 4.5 kN，起重机的旋转与固定部分共重G3 = 31 kN。尺寸如下图。设伸臂在起重机对称面内，且放在图示位置，试求车子不致翻倒的最大起吊分量Gmax。, 0F0321GGGGFFBA , 0FBM0)m 2m

8、 8 . 1 (m 2m 5 . 2)m 3m 5 . 2(12AFGGGGGGFA5 . 55 . 228 . 3121kN 7.52.525.5121GGq60GB60q60GmkN 2 .78960 sin360 coskN 10060 coskN 4 .31660 sin11FlFllFMMFPFFFFAAyAx 0360 sin60 cos , 0060 cos , 0060 sin , 011lFlFlFMMFMFPFFFFFFAAAyyAxxB60 (2)当平衡荷重当平衡荷重G3=180 kN时，时，求满载时轨道求满载时轨道A，B给起重机轮给起重机轮子的约束力？子的约束力？AFB

9、F kN 750m 2m 12m 2m 2m 60min321min3GGGGMBFkN 3500m 2m 2m 60max31max3GGGMAFAFBF81345 sin825GFGFGFAEyA045sin, 0045 cos, 002522, 0GFFFFFFlGlFFMEyAyExAxAE 02245 cos, 0lFlFlFFMEyKDBC823852GFGFGFDBExKDBFCyFCxF),2121(212qbpaMaFAy)2123(212qbMpaaFGFAyBy, 0)(FMB0222bqbaFaGMAy, 0yF0GFFByAy, 0 xF0qbFFBxAx)2321(

10、21)21(122qbpaMbqbaFbFAyAx)2121(212qbpaMbqbFFAxCx)2121(212paMqbaFFAyCy)2121(212qbpaMbqbFFAxBx, 0)(FMC02aFbqbbFAyAx, 0 xF0qbFFCxAx, 0yF0CyAyFF30FD 0m 4m 2, 00m 2m 4, 00, 0AyBByABxxFFMFFMFFFFkN 50ByAyBxFFF30A030 sin, 0030 cos, 0112FFFFFFAyyxkN 66.8kN 1021FF30FD1F030sin0030cos30cos041314FFFFFFFyxkN10kN1

11、034FF30D0025FFFxkN 66.85FD3F2F30D 0m 3m 1m 2, 00, 00, 0AxGEBGEByAxyAxxFFFFMFFFFFFFkN8kN90ByAyAxFFF0m 5 .1m 123m 5 .00060sin00m 1m 1230321AyEDEAyyAyEFFFMFFFFFFMFF1231 0 .4 k N1 .1 5 k N9 .8 1 k NFFF 如图平面桁架，求各杆内力。知铅垂力FC=4 kN，程度力FE=2 kN。0EAxFF，0 xF0CAyBFFF, 0yF03 aFaFaFBEC , 0FAM,kN 22AFFkN 4ACF，kN 2FE

12、FkN 2FCF045 cosAFACAxFFF，0 xF045 cosAFAyFF, 0yF, 0 xF045 cosFAFEFF, 0yF045 cosFAFCFF045 cosCECDCAFFF, 0 xF045 cosCECFCFFF, 0yF,kN 22CEFkN 2CDF0DCDBFF, 0 xF0DEF, 0yF,kN 3DBF0DEFkN 22BDFkN 22BEF045 cosBEBDFF, 0 xF, 0yF045 cosBEBFF 如图平面桁架，求如图平面桁架，求FEFE，CECE，CDCD杆内力。知铅垂力杆内力。知铅垂力FC = FC = 4 kN4 kN，程度力，程度力FE = 2 kNFE = 2 kN。0EAxFF,0 xF0CAyBFFF,0yF03aFaFaFBEC , 0FAM 5.作一截面m-m将三杆截断，取左部分为分别体，受力分析