1、Statically determinate structureStatically determinate structure0 xF0yF0M , Statically indeterminate beam Satically indeterminate frame Satically indeterminate composite Structures 確定確定超靜定次數(shù)超靜定次數(shù)FPFPFAxFAyMAX1靜定結(jié)構(gòu)靜定結(jié)構(gòu)AABBCCFPFPFAxFAyMAX1基本體系基本體系A(chǔ)ABBCCqqqAABBCCllX1X1b)基本體系基本體系a)一次超靜定結(jié)構(gòu)一次超靜定結(jié)構(gòu)Xj多余未知力多

2、余未知力基本體系沿多余未知力方向的位移應(yīng)與原結(jié)構(gòu)位基本體系沿多余未知力方向的位移應(yīng)與原結(jié)構(gòu)位移相同移相同F(xiàn)PFPFAxFAyMAX1基本體系基本體系A(chǔ)ABBCCqqqAABBCCllX1X1X11P2PnPFPAABBC1PX1ABX1=1d d1111 =d d11X1FPABCEIl/2l/2X1基本體系基本體系(1=B =0) =+XjXjABX1=1d d11FPABCEIl/2l/2X1基本體系基本體系(1=B =0) =FPAABBC1PX111 =d d11X1+單位力單位力d1d2dn,.,. 022110nnnnnnXXXddddnnSXSXSXSS22110=Z1R2112

3、34134PR2P12234R22R12R11R1PZ2PL2l2l1234EI=常數(shù)常數(shù)Z1Z2(a)(b)基本結(jié)構(gòu)基本結(jié)構(gòu)1234=Z1Z2R1=0=0PR2R1PR2P ,r11、r121121ZZ、1121ZZ、MPM1、M2 基本結(jié)構(gòu)在荷載等外因和結(jié)點(diǎn)位移的共同作用下，每一個(gè)附基本結(jié)構(gòu)在荷載等外因和結(jié)點(diǎn)位移的共同作用下，每一個(gè)附加聯(lián)系中的附加反力矩或反力都應(yīng)等于零加聯(lián)系中的附加反力矩或反力都應(yīng)等于零(靜力平衡條件靜力平衡條件)。r11Z1+ r12Z2+R1P=0r21Z1+ r22Z2+R2P=0P2211MZMZMMr11Z1+ r12Z2+R1P=0r21Z1+ r22Z2+R

4、2P=0R1=R11+R12+R1P=0R2=R21+R22+R2P=0力矩分配法力矩分配法ABCABC4m4m6mDFPBBABMMBAMBCMCB=10kNEIEI=10kNPF桿端彎矩M總MABBAMBCMCBM(主攻目標(biāo))ABCABCBPFPFDDMBBM =10100鎖住節(jié)點(diǎn)=10kN-101010=10kN不平衡力矩=BMFj=10ABMM固端彎矩FF=10=10FMBABCMF=0=0FMCB圖(kNm)PMMF怎么求？(問(wèn)題之一：)(即約束反力矩)分配系數(shù)BCBCM=(0.5)MB=5BAC傳遞系數(shù)()12CMABBA5=BMC傳遞彎矩MM分配彎矩=(0.5)MBA-2.5待分配力矩=BM=MB-5-5BBM =-10-5-5放松節(jié)點(diǎn)BBBCBACBAM圖(kNm)B10(問(wèn)題之二：怎么求？(問(wèn)題之三：C怎么求？ )分配系數(shù)BCBCM=(0.5)MB=5BAC傳遞系數(shù)()12CMABBA5=BMC傳遞彎矩MM分配彎矩=(0.5)MBA-2.5待分配力矩=BM=MB-5-5BBM =-10-5-5放松節(jié)點(diǎn)BBBCBACBAM圖(kNm)B10(問(wèn)題之二：怎么求？(問(wèn)題之三：C怎么求？ )彎矩分配系數(shù)彎矩分配系數(shù)For a unit rotation the moment in each member at joint A is j