結構難題!!!! - 土木
By Catherine
at 2010-01-19T00:59
at 2010-01-19T00:59
Table of Contents
※ 引述《fruitcake (水果蛋糕)》之銘言:
: 小弟 有一個大難題!!!
: 結構期末考
: 就靠這必考題過關
: 可是無奈天資愚笨
: 望版上大大高抬貴手
: 救救我
: 這個結構笨蛋
: 題目:
: http://ppt.cc/,@kV
: 拜託各位大大
: 謝謝
我這是傾角變位的解
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
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...
--
每個人 都戲茯隻菑v的未來 留下點六..
--
: 小弟 有一個大難題!!!
: 結構期末考
: 就靠這必考題過關
: 可是無奈天資愚笨
: 望版上大大高抬貴手
: 救救我
: 這個結構笨蛋
: 題目:
: http://ppt.cc/,@kV
: 拜託各位大大
: 謝謝
我這是傾角變位的解
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
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...
--
每個人 都戲茯隻菑v的未來 留下點六..
--
Tags:
土木
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