邊界元素法期中報告 第八次作業口頭報告 euler 樑單、三層勢能解法....
Post on 19-Dec-2015
237 Views
Preview:
TRANSCRIPT
邊界元素法期中報告第八次作業口頭報告
Euler 樑單、三層勢能解法
組員介紹
陳柏源 M93520010( 結構組 )
袁倫欽 D93520010( 水資組 ) 莊世璿 M93520059( 海工
組 )
大綱 Exact solution
1. 直接積分2. 力平衡 & 面積力矩法3. Direct BEM
Indirect BEM1. S:-1~2
2. S:2~3
3. S:0~1
Plot Indirect BEM Rank [A]
Problem
Exact solution
直接積分直接解控制方程式
d4uxdx4
0
vx d3uxdx3
C1
mx d2uxdx2
C1xC2
x duxdx
12C1x
2 C2x C3
ux 16C1x
3 12C2x
2 C3xC4
代入邊界條件求出係數
u0 016C103 1
2C202 C30C4 0
C4 0
0 012C102 C20 C3 0
C3 0
v1 F0C1 F0
m1 M0C11 C2 M0C2 M0 C1 M0 F0
代求出的係數回方程式
ux 16F0x
3 12M0 F0x2
力平衡 &面積力矩法0F
0M
+
-
+
力平衡Fy 0
F0 v0 0v0 F0M 0
m0 M0 F0 1 0m0 F0 M0利用面積力矩法, 彎矩圖面積
1 12
1 F0 1 M0
彎矩圖面積 重心位置 u
u1 12
1 F0 13
1 1 M0 12
assume 0 x 1U1, x 112
1 x3U0, x 1
12x3
;1, x 141 x2
0, x 14x2
;M1, x 121 x
M0, x 12x
;V1, x 12
V0, x 12
ux U1, xv1 1, xm1 M1, x1 V1, xu1U1, xv1 1, xm1 M1, x1 V1, xu1
112
1 x3 F0 141 x2 M0 1
21 x1
2F0 2 M0 1
2162 F0 3 M01
12x3 F0
14x2F0 M0
163 M0 F03 xx2
Direct BEM
uxUs, xvs s, xms Ms, xs Vs, xuss 1s 0
xUs, xvs s, xms Ms, xs Vs, xuss 1s 0
mxUms, xvs ms, xms Mms, xs Vms, xuss 1s 0
vxUvs, xvs vs, xms Mvs, xs Vvs, xuss 1s 0
U 112
s x3, x s
112
x s3, x s; 1
4s x2, x s
14x s2, x s
; M 12s x, x s
12x s, x s
; V 12, x s
12, x s
U 14
s x2, x s
14x s2, x s
; 12s x, x s
12x s, x s
; M 12, x s
12, x s
; V 0
Um 12s x, x s
12x s, x s
; m 12, x s
12, x s
; Mm 0 ; Vm 0
Uv 12, x s
12, x s
; v 0 ; Mv 0 ; Vv 0
d/ds
d/dx
u0 U1, 0v1 1, 0m1 M1, 01 V1, 0u1U0, 0v0 0, 0m0 M0, 00 V0, 0u0
112
F0 14M0
12
1 12u1... ...1
u1 U1, 1v1 1, 1m1 M1, 11 V1, 1u1U0, 1v0 0, 1m0 M0, 10 V0, 1u0
12
u1 112
v0 14m0... ...2
0 U1, 0v1 1, 0m1 M1, 01 V1, 0u1U0, 0v0 0, 0m0 M0, 00 V0, 0u014F0
12
M0 12
1... ...31 U1, 1v1 1, 1m1 M1, 11 V1, 1u1U0, 1v0 0, 1m0 M0, 10 V0, 1u0
12
1 14v0 1
2m0... ...4
1 12F0 2 M0
u1 162 F0 3M0
v0 F0m0 F0 M0
八個邊界；已知四個
assume 0 x 1U1, x 112
1 x3U0, x 1
12x3
;1, x 141 x2
0, x 14x2
;M1, x 121 x
M0, x 12x
;V1, x 12
V0, x 12
ux U1, xv1 1, xm1 M1, x1 V1, xu1U1, xv1 1, xm1 M1, x1 V1, xu1
112
1 x3 F0 141 x2 M0 1
21 x1
2F0 2 M0 1
2162 F0 3 M01
12x3 F0
14x2F0 M0
163 M0 F03 xx2
Indirect BEM
S: -1~2u0 U2, 02 M2, 02 U1, 00 M1, 0134
2 2 112
1 121... ...5
0 U2, 02 M2, 02 U1, 01 M1, 01 2 1
22 1
41 1
21... ...6
m1 Um2, 12 Mm2, 12 Um1, 11 Mm1, 11322 1... ...7
v1 Uv2, 12 Mv2, 12 Uv1, 11 Mv1, 11
122 1
21... ...8
5,6,7,8聯立0
34
2 2 112
1 121
0 2 122 1
41 1
21
M 322 1
F 122 1
21
1 2M 6F5
, 2 2M 4F5
, 1 8F 51M45
, 2 19M 48F45
assume 0 x 1ux U1, x1 M1, x1 U1, x0 M0, x0163M F3 xx3 O.K.
S: 2~3u0 0112
2 031 112
3 032 122 01 1
23 02 0
81 272 121 182 0 a0 0
142 021 1
43 022 1
21
122 0
41 92 21 22 0 bv1 F0
121
122 F0
1 2 F0 cm1 M0122 11
123 12 M0
1 22 M0 d
聯立abcd得1 2F0 M02 F0 M0
1 13F0
133
M0
2 56F0
116
M0
代回uxj1
2
Usj, xj j1
2
Msj, xj且令0 x 1得
ux 16F0x3
12M0 F0x2
O.K.
S: 0~1ux Us, xs Ms, xsu0 U1, 01 M1, 01 U0, 00 M0, 00
112
1 12
1... ...50 U1, 01 M1, 01 U0, 00 M0, 00
141 1
21 1
20... ...6
m1 Um1, 11 Mm1, 11 Um0, 10 Mm0, 10
12
0... ...7v1 Uv1, 11 Mv1, 11 Uv0, 10 Mv0, 10
121 1
20... ...8
5,6,7,8聯立0
112
1 12
10
141 1
21 1
20
M0 12
0F0
121 1
20
0 2 M0, 1 2F0 M0, 0 2F0 M03
,
1 F0 M03
assume 0 x 1ux U1, x1 M1, x1 U1, x0 M0, x0163M0 F03 xx2 O.K.
u0 112
1 12
1;0
141 1
21 1
20;
m1 12
0;v1 1
21 1
20;
代入ux,ux U1, x1 M1, x1 U1, x0 M0, x0
16F04 3 x x2 M04 3 x 2 x2 NG
Check x=0-,1+ (S 落入域內 )
與正解不同
相差兩處正副號
Plot indirect BEM
Plot Indirect BEM
Rank
A.1212
00F0M0
;
112
34
12
1
14
1 12
12
1 32
0 0
12
12
0 0
.1212
00F0M0
;
RankA Rank112
34
12
1
14
1 12
12
1 32
0 0
12
12
0 0
4
Rank [A] -1~2
Rank [A] 2~3
A.1212
00F0M0
;
8 27 12 18
4 9 2 2
1 1 0 0
1 2 0 0
.1212
00F0M0
;
RankA Rank8 27 12 18
4 9 2 2
1 1 0 0
1 2 0 0
4
Rank [A] 0~1
A.1212
00F0M0
;
0 1
120 1
2
0 14
12
12
12
12
0 0
12
0 0 0
.1212
00F0M0
;
RankA Rank0 1
120 1
2
0 14
12
12
12
12
0 0
12
0 0 0
4
The End
top related