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[傾角變位法]自由度`theta _B,Delta_C`
(一)相對勁度
`K_AB:K_BC:K=(2*720)/9:(2*200)/5:30=160:80:30(令80=k)`
`=>2k:k:30`
(二)固端彎矩
`M_(AB)^F=-(4*9^2)/12=-27(tf-m) ;M_(BA)^F=27(tf-m)`
`M_(BC)^F=-(3*48*5)/16=-45(tf-m)`
(三)側位移
`令R_(BC)=+R=(Delta_C)/5`
(四)列桿端彎矩
`M_(AB)=2k(theta _B)-27`
`M_(BA)=2k(2theta _B)+27`
`M_(BC)=k(1.5theta _B-1.5R)-45`
(五)力平衡方程式
`sumM_B=0=>k(5.5theta _B-1.5R)=18`
`sumF_y|_c=0=>k(1.5theta _B-1.5R)+75=5*(30)Delta_C=25(30)R`
`=>[(5.5theta _B-1.5R=0.225...........(1)),(1.5theta _B-10.875R=-0.9375.......(2))]`
`由(1)(2)聯立得`
`[(theta _B),(R)]=[(0.066938(rad)),(0.09544(rad))]`
將`theta _B,R`代入`M_(AB),M_(BA),M_(BC)`中
`[(M_(AB)=-16.3(tf-m)(逆時鐘)),(M_(BA)=48.42(tf-m)(順時鐘)),(M_(BC)=-48.42(tf-m)(逆時鐘)),(Delta_C=5*R=0.4772(m)(darr))]...............(Ans)`