1、STRUCTURAL MECHANICS结 构 力 学 7-4 位移法解有 侧 移 刚架用位移法计算有侧移刚架时，与无侧移刚架相比增加了新内容：(1)在基本未知量中，除了角位移外，还有线位移；(2)在杆件计算中，要考虑线位移的影响；(3)在建立基本方程时，要增加与结点线位移对应的平衡方程。DC 3 4Mi=2BA 14 6 3 44 1 2BM i i = + 2AB 12 6 3 44 1 2BM i i = 1、有侧移刚架简例 B0BM =0B A B CMM+=B = =3kN/m8m4m2ii iAB CDBC 3(2 ) BMi=BAMB BCMQBAF QCDF0 xF =0Q B
2、 A Q CDFF+= ABABMQABFQBAFBAM0AM =1 ( ) 64Q B A A B B AF M M= + DCQDCFQCDFDCM0DM = 14Q C D D CFM=4 3 4 2 0A B B A Q B AM M F+ + + =DC 2 23Mih=BA 113Mih=2、例题：试作图示刚架的弯矩图，忽略横梁的轴向变形。321/3 jPj jiF h= 0 xF = ( ) 0P Q A B Q CD Q E FF F F F + + =1 213QABFih=FPh1 h2h3I1 I2 I3AB D FCE111EIih= 22 2EIi h= 33 3EI
3、i h=FE 3 33Mih=2 223Q CDFih= 3 233QEFFih=FP AB D FCEQABFQCDFQEFFFP M 图3、例题：试作图示刚架的弯矩图。0BM =0B A B C B EM M M+ + =2BA 13 2 0 48B A BMi = + EBEBMQEBFQBEFBEMA B C DEF5m4m 4m4m2m20kN/m4I 4I5I3I 3I2BC 14 2 2 0 512B C B B C CM i i= + 2CB 12 4 2 0 512B C B B C CM i i= + + CD 3 C D CMi=BE 46B E B B EBEM i i l = EB 26B E B B EBEM i i l = CF 46C F C C FCFM i i l = FC 26C F C C FCFM i i l = MBA MBCMBEMCB MCDMCF0CM =0C B C D C FM M M+ + =QBEF QCFF0 xF = 0Q B E Q CFFF+=FCQFCFQCFFFCMCFM0EM =1 ()4Q B E B E E BF M M= +0FM =1 ()6Q C F C F F CF M M= +;BC= = =U R DISMISSEDCLASS IS OVER
