习题课2 静定平面桁架的内力计算

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1、习题课2静定平面桁架的内力计算一、找出桁架的零杆(1)FPFP000000008根零杆一、找出桁架的零杆(2)FP00000FP5根零杆FP(3)FPFP000000012根零杆0000FP(4)FPFP0000000000AA0012根零杆(5)CCa000aDDFPFPFP00FPaaAB2FPA0BFN1aaaaFPFP6根零杆由于荷载反对称，该桁架除下部水平链杆AB外，其余杆件受力反对称，故F=0。NCD(6)FFPP附属部分0000006根零杆(7)00FFPP000007根零杆二、用简捷方法求桁架指定杆轴力(1)E1Fa/2HG2A2DBa/CFPFPFPaa/2a/2a

2、Ⅱ解：E1FII-II左：a/2H∑MC=0G2A2DBa/Fa⋅+1.5Fa⋅=0NP1CⅠⅠFFNP1=−1.5(压)1.5FⅡFPFPFP1.5FPPaa/2a/2aI-I下：简单桁架∑MD=0FaFa+=/20F=−0.5FyP22yP515FF=−⋅=0.5−1.118F(压)NP2P21(2)3FP2F13Pd2ACDBFPdddd3FPⅡⅠ2F13Pd2FxA=2FPAFNCⅡDBⅠFFyA=FPFyB=2FPPF=FyCPdddd解：联合桁架I-I右：1)∑MC=0FdFdNP33+−=2FdP0FNP=F(拉)2)∑Fx=0FFNP=(拉)3)∑Fy=0FFyC=P

3、()↑3FPⅡⅠ2F13Pd2FxA=2FPAFNCⅡDBⅠFFyA=FPFyB=2FPPF=FyCPdddd整体平衡：1∑MB=0FyA=(3FdFP+⋅−P2d2FdFdP−P)=↑FP()2dII-II左：1∑MDN=0F1=−⋅=−(FdP)FP()压d∑Fy=0FFFFNPP2=−=3−2()P压(3)FPFPAE1aBF2Da/2Caa/2a/2aⅠFFPP0AE1-FP-FPaFPⅠB-FF020DPa/2FPCFN2FPaa/2a/2a联合桁架解：1)I-I右：∑MFEN=00=2)结点C：F=0N23)结点F：∑Fy=0FFFyPN11==2()FP拉(4)AFC1

4、G3mH2343mBDEFPFPFPFP4m4m4m4m6.67FPAⅡFC1ⅠG3mH236.67FP4FN3mBⅡDⅠEFPFPFPFP4FP4m4m4m4m解：简单桁架1)I-I右：22∑∑MFFDNP=01=()拉压FFFxNP=0=−()33FN22)结点E：∑F=0F=2FF=55F=F()拉23FPExx2PN22xP346FP6.67FPAⅡFC1ⅠG3mH236.67FP4FN3mBⅡDⅠEFPFPFPFP4FP4m4m4m4m3)II-II右：−1∑MF=0FN4=(4FFPP+=8)−2FP(压)6F4)结点D：N5FN3∑Fx=02FPD2FP/3434FPF

5、=FF=⋅=FFF=0x55PyPPN3343(5)2dCAB1FPFPFPdCddddⅠ复杂桁架02d0ABC11.5FFPFPFPP1.5Fd00PⅠC解：dddd1)结点C：结构与荷载均对称，两斜杆轴力为零。2)结点A：∑Fy=0FFNP2=()拉123)结点B：∑Fy=0Fy11=−=FFPN−FP()压2224)或取I-I左：∑Fy=0Fy11=−=0.5FFPN−FP(压)2(6)B13aACFPFPFPFPaaaaaB1联合桁架3aACFPFPFPFPaaaaa解：作如图封闭截面，取ABC部分为隔离体:−1∑MA=0FN1=(FaFaPP⋅+⋅=−2)0.6FP(拉)5

6、a(7)aCa1aABFPFPaaa联合桁架aCa1aA0BⅠⅠF解：FPFBx1PaaaI-I下：FPFPFFy1N112∑MAy=0F1=−⋅=−(FaP2)FP33a25525FN1=−=(FFPP)−()压1233(8)2FP1aCD2FP2aABaa2FP解：ⅠS11a1)I-I右：CD∑FS1=02FP2a22FF⋅+⋅34FAFNPPNB22Ⅰ2aa−⋅=20F3F3FPPP2复杂桁架FFNP=−()压2)结点B：F∑Fx=0FFFFxBC=PyBC=PFNBDNBCFF=2NBCPFBP∑FFFy=04NBD=−P3FP3)结点D：D2∑FFSN22=0−⋅=40FP

7、FN24F2PS2FF=22()拉NP2F=2F4)结点C：N1PC∑FFFx=0NCA=NCB∑Fy=0FFNP1=2()拉FFNCA=2PFFNCB=2P(9)1CFPaFPaFPaA2BDaaⅠ1CFP复杂桁架aFPaFPa3FA2PBDⅠ3FPaa3FP解：1I-I左：∑MD=0FN1=⋅=3FaFPP()拉3a1∑MCN=0F2=(3FPPP⋅−3a3Fa)=2F(拉)3a(10)CaD1aAFPBaa0.5FⅠCPaD1a0.75FAFPⅠPB