群共享13年7矩陣位移法

第六章矩陣位移法第六章矩陣位移法6.1

概

述矩陣位移法是以結(jié)構(gòu)位移為基本未知量，借助矩陣進(jìn)行分析，并用計(jì)算機(jī)解決各種桿系結(jié)構(gòu)受力、變形等計(jì)算的方法。理論基礎(chǔ)：位移法分析工具：矩陣

計(jì)算手段：計(jì)算機(jī)基本思想:化整為零

------

結(jié)構(gòu)離散化將結(jié)構(gòu)拆成桿件,桿件稱(chēng)作單元.單元的連接點(diǎn)稱(chēng)作結(jié)點(diǎn).631213565442e對(duì)單元和結(jié)點(diǎn)編碼.單元分析單元桿端力單元桿端位移集零為整結(jié)點(diǎn)外力結(jié)點(diǎn)外力------整體分析單元桿端力

單元桿端位移(桿端位移=結(jié)點(diǎn)位移)結(jié)點(diǎn)外力

結(jié)點(diǎn)位移基本未知量:結(jié)點(diǎn)位移6.2

矩陣位移法解連續(xù)梁P1P2P3i1

=

il1

=

li2

=

il2

=

l12123(1)(2)(3)1

2----單元編碼1,2,3

----結(jié)點(diǎn)編碼(1),(2),(3)

----結(jié)點(diǎn)位移編碼----整體編碼一.離散化結(jié)點(diǎn)位移逆時(shí)針為整,結(jié)點(diǎn)力逆時(shí)針為整.建立單元桿端力和單元桿端位移的關(guān)系.P1P2P3i1

=

il1

=

li2

=

il2

=

l12123(1)(2)(3)二.單元分析1,2----局部編碼eeF

eFe

{F}

=

2

1

----單元桿端力eieFe1Fe22de1

d12

2

1

----單元桿端位移ded

et12ioure

=

單元桿端力和單元桿端位移逆時(shí)針為正.單元分析的目的:ee

ieFe1Fe22de1

d12----單元桿端力二.單元分析1,2----局部編碼單元分析的目的:建立單元桿端力和單元桿端位移的關(guān)系.

ee

eF{F}

=

2

F1

2

1

----單元桿端位移ded

e5fgyudne

=

4ie1deFe1Fe22de2ie1e·d14iee2i2·de1F

e

=

4i

de

+

2i

de1

e

1

e

2F

e

=

2i

de

+

4i

deF

eF2

1

=

e e

1

2ie4i4ie

d2

2i

d

e2

e

1

e

2簡(jiǎn)記為{F}e

=

[k]e

miapjqke---單元?jiǎng)偠确匠谭Q(chēng)作單元?jiǎng)偠染仃?簡(jiǎn)稱(chēng)作單剛)其中[k]e二.單元分析單元?jiǎng)偠染仃囍性氐奈锢硪饬xee

ieFe1Fe22de1

d121dee4ieF1Fe22de11e2i·de4ie2e2i·de1F

e

=

4i

de

+

2i

de1

e

1

e

2F

e

=

2i

de

+

4i

deF

eF2

1

=

e e

1

2ie4i4ie

d2

2i

d

e2

e

1

e

2簡(jiǎn)記為{F}e

=

[k]e

ahhwr2te---單元?jiǎng)偠确匠谭Q(chēng)作單元?jiǎng)偠染仃?簡(jiǎn)稱(chēng)作單剛)其中[k]eeeke

keke

ke

[k]e

=

2ie

4ie

4i

2i

21 22

11 12

=

eijk---發(fā)生i端所需加的桿端力.eej

id

=1,d

=0

位移時(shí)在單元?jiǎng)偠染仃囆再|(zhì):對(duì)稱(chēng)矩陣三.整體分析整體分析的目的:建立結(jié)點(diǎn)力與結(jié)點(diǎn)位移的關(guān)系.1

11

1

21

2

31

3P

=

k

d

+

k

d

+

k

dP1P2P3i2

=

i1212

d1d23d3·d1i1

=

i=1k1131k=1·d212k=1·d3k2232k13k23kk33k21P2

=

k21d1

+

k22d2

+

k32d3P3

=

k31d1

+

k32d2

+

k33d3

33313

1

11

123

311

dkkk

d

k

kpP2

=

k21

k22

k23

d2

k32

P

簡(jiǎn)記為{P}=[k]{D}---結(jié)構(gòu)剛度方程[k]--結(jié)構(gòu)剛度矩陣(總剛)111k121k1k

=

1

k

=

k111

k11

21

21k31

=

0k

=

k1

k

=

k1

+

k

212

12

22

22

11k

=

k

232

2113k

=

0k

=

k

223

12k

=

k

233

221112k1k22112k122k111k

221k211

12

13k33

k31k22

k23

k32k

k

k

[k]=

k21簡(jiǎn)記為{P}=[k]{}D

---結(jié)構(gòu)剛度方程[k]--結(jié)構(gòu)剛度矩陣(總剛)k

=

1

k

=

k111

k11

21

21k31

=

0k

=

k1

k

=

k1

+

k

212

12

22

22

11k

=

k

232

2113k

=

0k

=

k

223

12k

=

k

233

22單元?jiǎng)偠染仃囍性氐奈锢硪饬x

ijk移為零位移時(shí)在i結(jié)點(diǎn)所需=1,其它結(jié)點(diǎn)位j---發(fā)生d加的結(jié)點(diǎn)力.結(jié)構(gòu)剛度矩陣性質(zhì):對(duì)稱(chēng)矩陣P1P2P3i2

=

i1212

d1d23d3·d1i1

=

i=111k=1·d212k=1·d3k2232k13k23kk3321k

k31111k121k1112k122k1112k122k111k

221k211

12

13k33

k31k22

k23

k32k

k

k

[k]=

k21P1P2P3i1

=

i i2

=

i1212

d1

d23d3簡(jiǎn)記為{P}=[k]{D}---結(jié)構(gòu)剛度方程[k]--結(jié)構(gòu)剛度矩陣(總剛)k

=

k1

k

=

k111

11

21

2131k

=

0k

=

k1

k

=

k1

+

k

212

12

22

22

1132

2113k

=

0k

=

k

223

122233k

=

k

2k

=

k

2單元?jiǎng)偠染仃囍性氐奈锢硪饬x

移為零位移時(shí)在i結(jié)點(diǎn)所需=1,其它結(jié)點(diǎn)位kij

---發(fā)生dj加的結(jié)點(diǎn)力.結(jié)構(gòu)剛度矩陣性質(zhì):對(duì)稱(chēng)矩陣總剛的形成方法---“對(duì)號(hào)入座”[k]=

21 22

11 12

1k1

k1k1[k]

=

1

21

2k1

1

1211k122

211

k11k121

223

21 22

11 12

2k2

k2k2[k]

=

2

31

22

3k

2

1

2k

+k21k22

2211

12k222k

030

1P1P2P3i

=

i1i

=

i21212

d1d23d3四.計(jì)算桿端力{P}=[k]{D}{F}e

=

[k]e

fpphttke計(jì)算結(jié)點(diǎn)位移計(jì)算桿端力四.計(jì)算桿端力{P}=[k]{D}{F}e

=

[k]e

lncpolhe計(jì)算結(jié)點(diǎn)位移計(jì)算桿端力6kN.m1i

=12i

=

23kN.m3kN.m例:

計(jì)算圖示梁,作彎矩圖解:12123(2)(3)804

4

2

0[k]=

2

1241[k]

=

21.離散化1

24

2

21

2

4 2

1

1

2[k]

=

42.計(jì)算總剛,總荷2

38

2

31

2

8

4

1

2

3

=-3-6{P}

{P}=[k]{D}

{

}

11/

24

D

=

-1/

6

3.解方程,求位移

-17/12(1)4.求桿端力

=

-6

2

4

-1/

6

-7

/

24

2-17/12{F}1

=

=

3

8 4

-1/

6

1/

22{F}

=

64

811/

247/21/23MM1-604

12244

8

dd3

=

0

0{D}=

0-3/

2

=

-604

-324 2-3/

21{F}

=

80

0440

0=82{F}

=2

0

0d

63N.m1i

=12i

=

23kN.mP312123(1)(2)(3)6k五.(零位移)邊界條件處理先處理法方法:

后處理法后處理法:

置0置1法乘大數(shù)法(1)置0置1法0

d2

=-30

0

1

3

P3

0-600

24

2120

3

1d

0

d2

=-30d1

1i

=12i

=

23kN.m作彎矩圖練習(xí):i2

=

23kN.mi1

=1作彎矩圖練習(xí):123(1)(2)(3)12

331

104

4

2P8

d2

12

d2

=-34d1

=d3

=

0

0

00

0

3

1

0120

1

dd2

=-30d1

0

0031dd2

=-1/

40d

P

d

-1

-1/

40

-1/

2=422

{F}1

=

4

=

-208

-18

4-1/

42{F}

=M1/24121

3

3

1

-608d

P

4

22

12