結構難題!!!! - 土木
By Daph Bay
at 2010-01-20T10:04
at 2010-01-20T10:04
Table of Contents
: 假設中點是C
: 2EI/L
: AC : BC = 6/8 : 4/8 = 3:2
: fMac = 0
: fMca = 0
: fMcb = PL/8 = 8*4/8 = -4
這邊不是 8*8/8=-8 嗎
: fMbc = 4
: Mac = 3*(2θa+θc-3Δ/L)+fMac θa=0
: =3(θc-3Δ/8)
: Mca = 3(2θc-3Δ/8)
: Mcb = 2*(2θc+θb+3Δ/8)+fMcb θb=0
: =2(2θc+3Δ/8)-4
: Mbc = 2(θc+3Δ/8)+4
: ΣMc =0 Mca+Mcb=0 ------ 1式
: 從ac構件得知
: 求得Ra
: 從cb構件
: 求得Rb
: 這裡反力 自己畫一下自由體圖
: 找一下
: Ra + Rb = 8 -------2式
: 解聯立
: 可以得到θc 跟 Δ
: 再回頭找Mac跟Mbc 就是Ma跟Mb
: 應該沒錯吧= =
: 別問我為什麼只會傾角
: 因為彎矩分配法學校沒教...
我自己用傾角算 Mab = -16.08(t-m) (後面就沒算了 我認為Mab對 後面大概都對)
我想知道一下正確答案是多少 有那個好心版友分享一下~~
這邊我對於會有Δ還是感到很奇怪 我看題目用感覺好像沒有Δ 但實際是有的
如果把固定端換成滾接+彎矩和鉸接+彎矩 這樣還是有Δ嗎
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土木
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