(1)求截面A的铅垂位移Av 截面A处添加一铅垂集中力F,刚架的应变能 2E So(qly)dy+5[ql--(ql+ F)x], +l' ay2)dy,+3[912+(qI-F)x,]'d 4V32%agy2-kx(-)dx1+2 lx2](--)dx 3l4 32E(向上) (2)求A截面水平位移A1 截面A处添加一水平面的集中力F,刚架的应变能 VE 2E (Jo[ql+F)y 1 dy+Jol(ql+F-g+ F)xi 'dx 2 +jo(qy2)dy2+5[ql+(+ F)x21 dx23 d,、Ol 2EⅠ {02h2dy1+∫2(q2-x1)-(-x1) +的2142+14dx}=3 (向右) 4EⅠ (3)求截面A的转角6 在截面A处,添加一集中力偶M4,刚架的应变能 (q4+12-(+Mxdx +l'cay2)dy2+olal )x2]2dx2} 62202、9、x)dx+b20为)7} 48EⅠ (4)求截面B的转角6B 截面B添加一集中力偶MB,刚架的应变能 r(alv) ' y,+nr (+M1)x2dx+(M-2)dy} (逆) 2EⅠ 解:(c) (1)截面A处的铅垂位移4 令作用于A处的集中力F=F1,刚架的应变能 Fx)dx+ (FI+ Fy)d  $ '$9 $ ) ¿ ¾ ½ ¯ ® ­ ³ ³               @ G     G >    TO\ \ TO TO ) [ [ (, 9 O O  ³ ³    O O T\ \ TO TO ) [ [              @ G      G >     G `  @      G >  @   ^ >               9  ³ ³      w w O O [ [ [ TO TO[ [ TO TO[ ) (, 9 '$ ) (, TO      $ '$+ $ ) ³ ³      O O ) [ [ TO TO ) \ \ TO ) O (, 9           @ G  ^ >  @ G >      ³ ³    O O ) [ [ TO T\ \ TO             @ G `      G >    ³ ³     w w O \ $ ) [ O [ [ TO TO\ \ TO ) (, 9         +      G  ^  G    ' (, TO TO TO[ [ [ O    G `              ³   $ T $ $ 0 $ ³ ³    O O $ [ [ O TO 0 TO\ \ TO (, 9             @ G  ^   G >     `            @ G       G >    [[ TO 0 T\ \ TO O O $  ³  ³    w w 0 $ $ $ 0 9 T  G `      G   ^                  ³ ³     O O [ O [ [ TO [ TO O [ [ TO TO (, (, TO     % T % % 0 % 9  `            G   @ G     G >    \ T\ [ [ 0 O TO 0 TO\ \ TO (, OO O % %  ¯ ® ­     ³³ ³ (, TO     F  $ '$9 $ ) ) ³ ³   O O ) O )\ \ (, ) [ [ (, 9         G     G