98結構技師結構學第四題

[study1]

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▼ 參考資料

`[矩陣位移法]`
`設定桿件ad=q1,ac=q2,ab=q3`
`[a]=[(1,0),(0,1),(-4/5,-3/5)]`
桿件勁度矩陣`[k]=((EA)/100)[(1/4,0,0),(0,1/3,0),(0,0,1/5)]=10((KN)/(cm))*[(1/4,0,0),(0,1/3,0),(0,0,1/5)]`
`[K]=[a]^T[k][a]=10((KN)/(cm))*[(0.378,0.096),(0.096,0.4053)]`
外力矩陣[R]
`[R]=[(0),(-100)](KN)`
`[K]{r}=[R]`
`=>10((KN)/(cm))*[(0.378,0.096),(0.096,0.4053)]*[(r1),(r2)]=[(0),(-100)](KN)`
`=>[(r1),(r2)]=[(6.67(->)),(-26.25(darr))](cm)=[(0.0667(->)),(-0.2625(darr))](m)`
求各桿件內力
`[S]=[k][a]{r}=[(q1),(q2),(q3)]=[(16.675(KN)),(87.5(KN)),(20.83(KN))]皆為拉力`

[study1]

`[最小功法]`
`取ab桿為贅力R,令外力100(KN)為P`
由靜力節點法求得
`S_(ad)=(4R)/5 ,S_(aC)=P-(4R)/5 ,S_(ab)=R`
其偏微分`(delS(R))/(delR) `
`(delS_(ad))/(delR)=(4)/5 ,(delS_(ac))/(delR)=(-3)/5 ,(delS_(ab))/(delR)=1`
`L_(ad)=4(m),L_(ac)=3(m),L_(ab)=5(m),`
根據最小功法標準式`sum(S*(delS_(i)(R))/(delR)*L_(i))/(E_(i)A_(i))`,因為`E_(i)A_(i)=constant`
`:.(sumS*(delS_(i)(R))/(delR)*L_(i))/(EA)`
`而(sumS*(delS_(i)(R))/(delR)*L_(i))=(216R)/(25)-(9P)/5`
代入最小功法標準式
`sum(S*(delS_(i)(R))/(delR)*L_(i))/(EA)=sum(S*(delS_(i)(R))/(delR)*L_(i))=0`
`rArr R=(45P)/216=20.83(KN)`
將R代入`S_(ad),S_(ac)S_(ab)`
`S_(ad)=16.67,(KN)S_(ac)=87.5(KN),S_(ab)=20.83(KN)`
求`d_(1),d_(2)`
`d_(1)=delta _(ad)=(N_(ad)*L_(ad))/(EA)=(16.67*4)/1000=0.0667(m)(rarr)`
`d_(2)=delta _(ac)=(N_(ac)*L_(ac))/(EA)=(87.5*3)/1000=0.2625(m)(darr)`