: \" P# m. L' MTwo metallic beams a and c are fixed to a block as shown below. Both beam a and beam c remain perpendicular ) m9 x$ b) q/ w2 J6 d8 L0 Xto block b at the connecting points. Beam a is attached to a wall and it remains perpendicular to the wall at the , e4 D; W g, ^$ B8 q$ S; Z' \0 sother end. Beam c is passed through a hole in the wall. Direction of the hole is perpendicular to the wall surface. t, Y' ~7 l/ [# D$ g2 E4 g0 kRelated dimensions are shown on the diagram. The diagram is in millimeter. Beams a and c have a circular ! n0 b6 l( n& h6 x% o5 tcross-section with a diameter of 0.7mm. The Young’s modulus is 70 GPa. Both beam a and beam c are initially & l. L4 _! w( U. h2 ]0 R# _) b& p
straight. 9 R& v: W5 O+ dNeglect gravity. Assume a perfect linear relationship between stress and strain for beams a and c . Neglect axial " ~# D7 J5 l# r9 r! Felongation or compression of beams a and c . 9 A4 p: K1 P( {' m1 M% Y. g# |Using elliptic integrals, derive relevant entities to predict the final shapes of beams a and c when beam c is pulled % U1 X& v) z- g
for 10 mm in the indicated direction. # ?. Y+ U, J/ g8 e
Use Matlab to implement your derivations and plot the deflected shapes of beams a and c in a figure. You should 8 d3 }2 D! |' H
also plot block b at the desired position. Please make sure your plot has a correct X-Y scale so that the figure / N, W9 c1 I- }& U2 ]- Zlooks realistic. $ O* a& d" z' N; D) T. q# g
Please also compare how close the deflected shapes of beams a and c are to circular arcs, by drawing circular arcs 9 W2 c f& T3 U& ^
which pass through the two ends of beams a and c . The circular arcs should also be perpendicular to the wall ; T5 M" \# q/ D& W2 {. V7 x6 V, x( ^
surface at one end. , O1 e. Z% u* @$ M- f+ f[attach]306825[/attach]作者: 潑墨 時(shí)間: 2013-12-19 19:25
大概有10個(gè)以上的未知數(shù)和方程,，但是含有數(shù)個(gè)超越方程,，用matlab做仿真一直沒(méi)出來(lái)/ g, d- \( f2 U' X4 E; F. s
有大俠有好的方法么。,。,。,。。作者: 茉莉素馨 時(shí)間: 2013-12-19 20:42
這個(gè)沒(méi)有人會(huì)回的,，樓主放棄吧作者: wolalala12 時(shí)間: 2013-12-19 21:08
茉莉素馨 發(fā)表于 2013-12-19 20:42 6 Y: S; E0 Z9 N. j. I! ^( Q0 q
這個(gè)沒(méi)有人會(huì)回的,，樓主放棄吧
5 E9 C9 ?% o+ C1 G
同馬克0 Q5 S/ t& t8 _# o3 ], w 作者: 成形極限 時(shí)間: 2013-12-19 23:11
樓主這是在國(guó)外深造�,。�作者: 成形極限 時(shí)間: 2013-12-19 23:39
老實(shí)說(shuō),，我不會(huì)做,，看了下題目，估計(jì)了下,，估摸著是這種形狀,，哈哈5 R# S1 Q$ x: d5 t& \* p
8 p( m: m' q& d
應(yīng)該B的重力影響比較大吧) {/ e% H! [/ N I
' e% }. h0 t' \! J: J) u, A% T
[attach]306852[/attach]作者: liushaobo1989 時(shí)間: 2013-12-20 16:38
求變形嗎？可以使用有限元,？作者: 16443 時(shí)間: 2013-12-23 22:38
B看成剛性體! K0 j4 I; }! N2 z, |1 ^
或略a和c的軸向變形 5 j. ^; v0 }* K; ?. ?( C+ c$ N根據(jù)撓度方程聯(lián)列a和c的變形方程
6 Y: S; E0 Z9 N. j. I! ^( Q0 q