100鐵路特考高員級結構學參考答案
發表於 : 2011 6月 27 (週一) 6:04 pm一
圖一(a)不穩定,圖一(b)不穩定
二
B點位移`Delta_B=(PL^3)/(6EI)(darr),`
相對轉角`theta_(B,r)=(PL^2)/(2EI)`
三
`theta_B=(5PL^2)/(62EI)`順時針
`theta_C=(9PL^2)/(62EI)`順時針
`Delta_(C,h)=(73PL^3)/(372EI)(rarr)`
`N_(AB)=N_(BC)=(21P)/(31)`
`N_(BD)=(-46P)/(31)`
`N_(CE)=0`
`V_(AB)=-V_(BA)=(15P)/(31)(rarr)`
`V_(CB)=-V_(CB)=P(larr)`
`V_(BD)=-V_(DB)=(15P)/(62)(darr)`
`V_(CE)=-V_(EC)=(27P)/(62)(darr)`
`M_(AB)=(5PL)/(32)`
`M_(BA)=(10PL)/(31)`
`M_(BD)=(15PL)/(62)`
`M_(BC)=(-35PL)/(62)`
`M_(CB)=(-27PL)/(62)=-M_(CE)`
四
[K]=`(EI)/L*{(12,2,2,0,6/L),(2,8,0,2,6/L),(2,0,4,0,0),(0,2,0,4,0),(6/L,6/L,0,0,12/(L^2))]`
圖一(a)不穩定,圖一(b)不穩定
二
B點位移`Delta_B=(PL^3)/(6EI)(darr),`
相對轉角`theta_(B,r)=(PL^2)/(2EI)`
三
`theta_B=(5PL^2)/(62EI)`順時針
`theta_C=(9PL^2)/(62EI)`順時針
`Delta_(C,h)=(73PL^3)/(372EI)(rarr)`
`N_(AB)=N_(BC)=(21P)/(31)`
`N_(BD)=(-46P)/(31)`
`N_(CE)=0`
`V_(AB)=-V_(BA)=(15P)/(31)(rarr)`
`V_(CB)=-V_(CB)=P(larr)`
`V_(BD)=-V_(DB)=(15P)/(62)(darr)`
`V_(CE)=-V_(EC)=(27P)/(62)(darr)`
`M_(AB)=(5PL)/(32)`
`M_(BA)=(10PL)/(31)`
`M_(BD)=(15PL)/(62)`
`M_(BC)=(-35PL)/(62)`
`M_(CB)=(-27PL)/(62)=-M_(CE)`
四
[K]=`(EI)/L*{(12,2,2,0,6/L),(2,8,0,2,6/L),(2,0,4,0,0),(0,2,0,4,0),(6/L,6/L,0,0,12/(L^2))]`