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