土木人

[傾角變位法]
設三個自由度`theta_b.Delta_b(darr).Delta _c(darr)(Delta_b>Delta_c)`
`1.相對勁度:K_(ab):K_1:K_2=(2EI)/L:(2EI)/L^3:(EI)/L^3=k:(k)/L^2:(k)/(2L^2)`
`2.相對變位:令R_(ab)=Delta_b/L=+R.R_(bc)=(Delta_b-Delta_c)/L=(R-R_2)=-(R-R_2)`
`3.列出彎矩方程式:`
`M_(ab)=k(theta_b-3R)`
`M_(ba)=k(2theta_b-3R)`
`M_(bC)=(2EI=oo)/L(1.5theta_b+1.5(R-R_2))`
`M_(bC)=有界:.theta_b+R-R_2=0............(1)`
`取ab桿件力平衡得到V_a=(k(3theta_b-6R))/L(darr).M_(ab)=(順時針)`
`sumF_(y(整體))=0=>k(3theta_b-7R-0.5R_2)=-PL........(2)`
`sumM_(a點固定端)=0=>k(theta_b-4R-R_2)=-PL...........(3)`
`(1).(2).(3)式聯立=>得[(ktheta_b=(PL)/5),(kR=(PL)/5),(kR_2=(2PL)/5)]=>[(theta_b=(PL^2)/(10EI)(順時針)),(Delta _b=(PL^3)/(10EI)(darr)),(Delta _c=(PL^3)/(5EI)(darr))]`

不好意思
本題取`Delta_b,Delta_c`兩自由度即可
若是位法不好解,則採力法較佳
以上共參考(跪拜禮new)

[最小功法]
這是不錯的選擇 由其搭配體積積分法也算蠻快的在此不多加贅述
,而改用積分方式計算之
`取b,c點彈簧力分別為贅力R,Q`
`M(x)_(bc)=Qx``=>(delM(x)_(bc))/(delQ)=x
`M(x)_(ab)=(R-P)x+Q(L+x)``=>(delM(x)_(bc))/(delQ)=(L+x)`
`F_b=R``=>(delF_b)/(delQ)=0`
`F_c=-Q``=>(delF_c)/(delQ)=-1`
`由(delU)/delQ=0`
`int(Qx^2)/(EI=oo)+int((R-P)x(L+x)+Q(L+x)^2)/(EI)+(R*0)/k_1+(Q*1)/k_2=0`
`=>5/6R+10/3Q=5/6P...........(1)`
`(delM(x)_(bc))/(delR)=0``=>(delF_b)/(delR)=-1`
`(delM(x)_(ab))/(delR)=x``=>(delF_c)/(delR)=0`
`由(delU)/delR=0`
`int((R-P)x^2)/(EI)+int(Q(L+x)*x)/(EI)+(R*1)/k_1+0=0`
`=>5/6R+5/6Q=5/3P...........(2)`
`(2)-(1)式得`
` [(Q),(R)]=[(1/5P),(1/5P)]=>[(Delta_c),(Delta_b)]=[(Q/k_2),(R/k_1)]=[((PL^3)/(5EI)(darr)),((PL^3)/(10EI)(darr))]`
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[矩陣位移法]
`設定自由度:b點垂直為r_1,c點垂直為r_2`
`ab桿內力為q_1,q_2,彈簧力b點內力為f_1,f_2`
求`[a]`矩陣
`當r_1=1時[a_1]=[(-1)/L ,(-2)/L ,-1, 0]^T`
`當r_2=1時[a_2]=[0 ,(1)/L ,0 ,-1]^T`
`[a]=[((-1)/L,0),((-2)/L,(1)/L),(-1,0),(0,-1)]`
`[k]=[((4EI)/L,(2EI)/L,0,0),((2EI)/L,(4EI)/L,0,0),(0,0,(2EI)/L^3,0),(0,0,0,(EI)/L^3)]`
`[K]=[a]^T[k][a]=[((-1)/L ,(-2)/L ,-1, 0),(0 ,(1)/L ,0 ,-1)]*[((4EI)/L,(2EI)/L,0,0),((2EI)/L,(4EI)/L,0,0),(0,0,(2EI)/L^3,0),(0,0,0,(EI)/L^3)]*[((-1)/L,0),((-2)/L,(1)/L),(-1,0),(0,-1)]`
`[K]=[((8EI)/L^3+(20EI)/L^3+(2EI)/L^3,(-10EI)/L^3),((-10EI)/L^3,(5EI)/L^3)]=(EI)/L^3[(30,-10),(-10,5)]`
求外力矩陣`[R]`
`[R]=[(P),(0)]`
寫成`[K]{r}=[R]`
`(EI)/L^3[(30,-10),(-10,5)]*[(r_1),(r_2)]=[(P),(0)]`
`[(r_1),(r_2)]=[(P/10),(P/5)]*L^3/(EI)=[((PL^3)/(10EI)(darr)),((PL^3)/(5EI)(darr))]`

此題亦可使用圖解勁度法!!