四.位移分量: ln=[-(1+)-+Cr(1-) E u= Fr-I sin 6+Kcos e 若适当给定约束条件,无刚性位移 e leso, ue=z2=0>F=k=l=0 (1+)-+-(1-)r E 五.特例 1.只受内压4≠0,q6=0四.位移分量: 五. 特例 = − + = − + + − sin cos [ (1 ) (1 )] 1 u Fr I K Cr r A E ur 若适当给定约束条件,无刚性位移 u | =0 ,u | = / 2 = 0 → F = K = I = 0 = = − + + − 0 (1 ) 1 (1 ) u r E C E r A ur 1.只受内压 qa 0,qb = 0