06 Matrix Beam

8/6/2019 06 Matrix Beam

1/148

1

! Development: The Slope-Deflection

Equations

! Stiffness Matrix

! General Procedures

! Internal Hinges

! Temperature Effects

! Force & Displacement Transformation! Skew Roller Support

BEAM ANALYSIS USING THE STIFFNESS

METHOD


2/148

2

Slope Deflection Equations

settlement = j

Pi j kw

Cj

Mij Mjiw

P

j

i

i j


3/148

3

Degrees of Freedom

L

A B

M

1 DOF:

P

A

B

C 2 DOF: ,


4/148

4

L

A B

1

Stiffness Definition

kBAkAA

L

EIkAA

4=

LEIkBA 2=


5/148

5

L

A B1

kBBkAB

L

EIkBB

4=

LEIkAB 2=


6/148

6

Fixed-End Forces

Fixed-End Forces: Loads

P

L/2 L/2

L

w

L

8

PL

8

PL

2

P

2

P

12

2wL

12

2wL

2

wL

2

wL


7/148

7

General Case

settlement = j

Pi j kw

Cj

Mij Mjiw

P

j

i

i j


8/148

8

wP

settlement = j

(MFij) (MFji)

(MFij)Load (MFji)Load

+

Mij Mji

i

j

+

i jwP

MijMji

settlement = jj

i

=+ ji L

EI

L

EI

24

jiL

EI

L

EI

42

+=

,)()()2

()4

( LoadijF

ijF

jiij MML

EI

L

EIM +++= Loadji

Fji

F

jiji MML

EI

L

EIM )()()

4()

2( +++=

L


9/148

9

Mji Mjk

Pi j kwCj

Mji Mjk

Cj

j

Equilibrium Equations

0:0 =+=+ jjkjij CMMM


10/148

10

+

1

1

i jMij Mji

i

j

L

EIkii

4=

L

EIkji

2=

L

EIkij

2=

L

EIkjj

4=

i

j

Stiffness Coefficients

L


11/148

11

[ ]

=

jjji

ijii

kk

kkk

Stiffness Matrix

)()2

()4

( ijF

jiij ML

EI

L

EIM ++=

)()4

()2

( jiF

jiji M

L

EI

L

EIM ++=

+

=

F

ji

F

ij

j

iI

ji

ij

M

M

LEILEI

LEILEI

M

M

)/4()/2(

)/2()/4(

Matrix Formulation


12/148

12

[D] = [K]-1([Q] - [FEM])

Displacement

matrix

Stiffness matrix

Force matrixw

P(MFij)Load (MFji)Load

+

+

i jwP

MijMji

j

i

j

Mij Mji

i

j

(MFij) (MFji)

Fixed-end moment

matrix

][]][[][ FEMKM +=

]][[])[]([ KFEMM =

][][][][1 FEMMK =

L


13/148

13

L

Real beam

Conjugate beam

Stiffness Coefficients Derivation

MjMi

L/3

i

MM ji +

I

Mj

I

Mi

I

LMj

2

I

LMi

2

)1(2

0)3

2)(

2()

3)(

2(:0'

=

=+=+

ji

jii

MM

L

EI

LML

EI

LMM

)2(0)2

()2

(:0 =+=+ILM

ILMF jiiy

ij

ii

EI

M

L

EIM

andFrom

)

2

(

)4

(

);2()1(

=

=

MM ji +

i j


14/148

14

Derivation of Fixed-End Moment

Real beam

8,0

162

22:0

2

PLMI

PLI

MLI

MLFy ==+=+

P

M

MI

M

Conjugate beam

A

I

M

B

L

P

A B

I

M

I

ML

2

I

M

I

ML

2

IPL16

2

IPL4 I

PL16

2

Point load


15/148

15

L

P

841616

PLPLPLPL

=+

+

8

PL

8

PL

2

P

2

PP/2

P/2

-PL/8 -PL/8

PL/8

-PL/16-PL/8

-

PL/4

+

-PL/16-PL/8

-


16/148

16

Uniform load

L

w

A B

w

M

M

Real beam Conjugate beam

A

I

M

I

M

B

12,0

242

22:0

23

wLMI

wLI

MLI

MLFy ==+=+

IwL8

2

I

wL

24

3

I

wL

24

3

I

M

I

ML

2

I

M

I

ML

2


17/148

17

Settlements

M

M

L

MjMi = Mj

L

MM ji +MM ji +

Real beam

26LEIM =

Conjugate beam

I

M

A B

I

M

,0)3

2)(

2()

3)(

2(:0 =+=+

L

I

MLL

I

MLMB

I

M

I

ML

2

I

M

I

ML

2


18/148

18

Typical Problem

0

0

0

0

A C

B

P1P2

L1 L2

wCB

80

24 11

11

LP

L

EI

L

EIM BAAB +++=

8

042 11

11

LP

L

EI

L

EIM BABA ++=

1280

242

222

22

wLLP

L

EI

L

EIM CBBC ++++=

1280

422

222

22

wLLP

L

EI

L

EI

M CBCB

+++=

P

8

PL

8

PL

L

w12

2wL

12

2wL

L


19/148

19

MBA MBC

A C

B

P1P2

L1 L2

wCB

B

CB

80

42 11

11

LP

L

EI

L

EIM BABA ++=

128

024

2

222

22

wLLP

L

EI

L

EIM CBBC ++++=

BBCBABB forSolveMMCM ==+ 0:0


20/148

20

A C

B

P1P2

L1 L2

wCB

Substitute B

in MAB

, MBA

, MBC

, MCB

MABMBA

MBCMCB

0

0

0

0

80

24 11

11

LP

L

EI

L

EIM BAAB +++=

80

42 11

11

LP

L

EI

L

EIM BABA ++=

1280

242

222

22

wLLP

L

EI

L

EIM CBBC ++++=

1280

422

222

22

wLLP

L

EI

L

EIM CBCB

+++=


21/148

21

A B

P1

MAB

MBA

L1

A CB

P1P2

L1 L2

wCB

ByR CyAy ByL

Ay Cy

MABMBA

MBCMCB

By = ByL +ByR

B CP2

MBC

MCB

L2


22/148

22

2EI EI

Stiffness Matrix

Node and Member Identification

Global and Member Coordinates

Degrees of Freedom

Known degrees of freedomD4,D5, D6, D7, D8 andD9 Unknown degrees of freedomD1, D2 andD3

1 21 2 31

2 3

7

8

9

4

5

6

1

2 3

7

8

9

4

5

6


23/148

23

i j

E, I, A, L

Beam-Member Stiffness Matrix

d4 = 1

AE/LAE/L AE/L AE/L

6

4

5

1

2

3

[k] =

1 2 3 4 5 6

3

2

1

4

6

5

d1 = 1

k11= = k44

0

0

0

0

0

0

0

0

-AE/LAE/L

-AE/L AE/L

k41 k14


24/148

24

d2 = 1

[k] =

1 2 3 4 5 6

3

2

1

4

6

5

0

0

0

0

0

0

0

0

-AE/LAE/L

-AE/L AE/L

6EI/L2

6EI/L2

d5

= 1

6EI/L2

6EI/L2

12EI/L312EI/L3 12EI/L3

12EI/L3

6EI/L2

12EI/L3

- 6EI/L2

- 12EI/L3

0

0

6EI/L2

-12EI/L3

0

0

- 6EI/L2

12EI/L3

k22= = k55

= k52

k62 =

= k32= k65

= k25

= k35

i j

E, I, A, L6

4

5

1

2

3


25/148

25

[k] =

1 2 3 4 5 6

3

2

1

4

6

5

0

0

0

0

0

0

0

0

-AE/LAE/L

-AE/L AE/L

6EI/L2

12EI/L3

- 6EI/L2

- 12EI/L3

0

0

6EI/L2

-12EI/L3

0

0

- 6EI/L2

12EI/L3

4EI/L

6EI/L2 6EI/L2

2EI/L

0

0

2EI/L

-6EI/L2

0

0

-6EI/L2

4EI/L

d6 = 1

d3 = 1 2EI/L4EI/L

6EI/L26EI/L26EI/L2

4EI/L 2EI/L

6EI/L2

k33=

k23= = k53

= k63 = k36

= k56k26=

= k66

i j

E, I, A, L6

4

5

1

2

3


26/148


27/148

27

+

=

F

j

F

yi

F

xj

F

i

F

yi

F

xi

j

j

j

i

i

i

j

yj

xj

i

yj

xi

M

FF

M

F

F

LEILEILEILEI

LEILEILEILEILAELAE

LEILEILEILEI

LEILEILEILEI

LAELAE

M

FF

M

F

F

/4/60/2/60

/6/120/6/12000/00/

/2/60/4/60

/6/120/6/120

00/00/

22

2323

22

2323

F

jjjjiiij

F

yjjjjiiiyj

F

xijjjiiixj

F

ijjjiiixi

F

yijjjiiiyi

F

xijjjiiixi

MLEILEILEILEIM

FLEILEILEIF

FLAELAEFMLEILEILEILEIM

FLEILEILEILEIF

FLAELAEF

)/4()/6()0()/2()/6()0(

)/6()0()0()/6()/12()0(

)0()0()/()0()0()/()/2()/6()0()/4()/6()0(

)/6()/12()0()/6()/12()0(

)0()0()/()0()0()/(

22

223

22

2323

=

=

==

+=

+++++=

[q] = [k][d] + [qF]

Fixed-end force matrix

End-force matrix

Stiffness matrix

Displacement matrix


28/148

28

6x6 Stiffness Matrix

[ ]

=

LEILEILEILEI

LEILEILEILEI

LAELAE

LEILEILEILEI

LEILEILEILEI

LAELAE

k

/4/60/2/60

/6/120/6/120

00/00/

/2/60/4/60

/6/120/6/120

00/00/

22

2323

22

2323

66

i j ji ji

Mi

Vj

Mj

Vi

Nj

Ni


[ ]

=

LEILEILEILEI

LEILEILEILEI

LEILEILEILEILEILEILEILEI

k

/4/6/2/6

/6/12/6/12

/2/6/4/6/6/12/6/12

22

2323

22

2323

44

i j ji

Mi

Vj

Mj

Vi


29/148

29

Comment:- When use 4x4 stiffness matrix, specify settlement.

- When use 2x2 stiffness matrix, fixed-end forces must be included.


i jMiMj

[ ]

=

LEILEI

LEILEIk

/4/2

/2/422


30/148

30

2131

2

1

2

3

4

5

6

General Procedures: Application of the Stiffness Method for Beam Analysis

wP P2y

M2M1

Local

1

2

3

4

1

1

2

3

4

2

Global


31/148

31

Member

1

2

3

4

1

3

4

5

6

2

1

2

3

4

5

6

312

1 2Global

wP P2y

M2M1

Local

1

2

3

4

1

1

2

3

4

2


32/148

32

(q F4 )2

(qF

3 )2

(q F1)2

(qF

2)2

w

2

P

(q F4)1

(qF

3)1

(q F2)1

(q F1 )1

1[FEF]

1

2

3

4

5

6

312

1 2Global

Local

1

2

3

4

1

1

2

3

4

2

wP P2y

M2M1

2 4 6


33/148

33

[q] = [T]T[q]

1

2

3

4

5

6

31

21 2Global

Local

1

2

3

4

1

[k] = [T]T[q] [T]

=

'4

'3

'2

'1

4

3

2

1

10000100

0010

0001

qq

q

q

qq

q

q

1 42 3

1

2

34

1

2 4 6


34/148

34

[q] = [T]T[q]

1

2

3

4

5

6

31

21 2

Global

Local

1

2

3

4

2

[k] = [T]T[q] [T]

=

'4

'3

'2

'1

6

5

4

3

10000100

0010

0001

qq

q

q

qq

q

q

1 42 3

3

4

56

2

2 4 6


35/148

35

Stiffness Matrix:

21

31

2

1

2

3

4

5

6

[k]1 =

1

2

3

4

1 2 3 4

[k]2 =

3

4

5

6

3 4 5 6

Member 1

[K] =

1

23

4

5

6

1 2 3 4 5 6

Member 2

Node 2

1 2


36/148

36

+

=

F

F

F

F

F

F

Q

QQ

Q

Q

Q

D

D

D

D

D

D

Q

Q

Q

Q

Q

Q

6

5

1

4

3

2

6

5

1

4

3

2

6

5

1

4

3

2 2

3

4

15

6

5 612 3 4

0

0

0

M1z

-P2y

M3z

KBA

KAB

KBB

KAA

2131

2

1

2

3

4

5

6

Reaction

Qu

Joint Load

Qk

Du=Dunknown

Dk= D

known

QFB

QFA

[ ] [ ][ ] FAuAAk QDKQ +=

[ ] [ ] [ ] [ ])(1 FAkAAu QQKD +=

[ ] [ ][ ]F

qdkqForceMember +=:

Global: 2 4 62 4 6


37/148

37

+

6

5

4

3

2

1

D

D

D

DD

D

Global:

21 3

1

2

1

2

3

4

5

6

12

2

3

4

51

6

(q F6 )2

(q F5 )2

(qF4)2

(qF3)2

w

2

P

(qF4)1

(qF3)1

(qF2)1

(qF1 )1

1[FEF]

=

=

=

24

23

12

MQ

PQ

MQ

y

4

3

2

D

D

D

+F

F

F

Q

Q

Q

4

3

2

2 3 4

Node 2

=

2

3

4

Member 1

1

23

4

5

6

1 2 3 4 5 6

Member 2

Node 2=

6

5

4

3

2

1

Q

Q

Q

QQ

Q

+=

+=

F

F

F

4

F

4

F

F

3

F

3

F

F

F

Q

Q

qqQ

qqQQ

Q

6

5

214

213

2

1

)()(

)()(M1-P2y

M2

0

0

0

2 4 6


38/148

38

12

31

21

2

3

4

5

6

4

3

2

D

D

D

2 3 4

Node 2

=

2

3

4

=

=

=

F

F

F

Q

Q

Q

MQ

PQ

MQ

4

3

2

24

2y3

12

2 4 6


39/148

39

Member 1:

21 3

1

21 3 5

P

qF4

qF3

qF2

qF1

1

[FEF]

4

3

2

1

q

qq

q

=

==

=

44

33

22

1 0

Dd

DdDd

d

+

F

F

F

F

q

qq

q

4

3

2

11

23

4

1 2 3 4

= 1k

2 4 6


40/148

40

21 3

1

21 3 5

Member 2:

qF6

qF5

qF4qF3

w

2

[FEM]

6

5

4

3

q

qq

q

=

==

=

0

0

6

5

44

33

d

dDd

Dd

+

F

F

F

F

q

qq

q

6

5

4

31

2

3

4

1 2 3 4

= 1k


41/148

41

A

CB

9 m 3 m

10 kN

1 kN/m

1.5 m

Example 1

For the beam shown, use the stiffness method to:

(a) Determine the deflection and rotation atB.

(b) Determine all the reactions at supports.

(c) Draw the quantitative shear and bending moment diagrams.

10 kN


42/148

42

21Global

123

1Members

23

2

12

A

CB

9 m 3 m

10 kN

1 kN/m

1.5 m

wL2/12 = 6.75wL2/12 = 6.75 PL/8 = 3.75PL/8 = 3.75

10 kN

1.5 m1.5 m

2

1 kN/m

9 m 1[FEF]

123 123


43/148

43

4/3

2/3

2/3

21

9 m 3 m

[k]1= EI4/9

2/9

3

4/9

2/92

3

2[k]2 = EI

4/3

4/3

2/3

2/3

2 12

1

[K] = EI

2 1

2

1

(4/9)+(4/3)

21

Stiffness Matrix:i j

MiMj

[ ]

=

LEILEI

LEILEIk

/4/2

/2/422

10 kN 123


44/148

44

A C

B

1 kN/m

Equilibriumequations: MCB = 0

MBA + MBC= 0

Global Equilibrium: [Q] = [K][D] + [QF]

=2.423/EI

0.779/EI

C

B

9 m 1.5 m1.5 m

6.756.75

+-3.75

-6.75 + 3.75 = -3MBA+ MBC= 0

MCB= 01

2

C

B

4/3

2/3

2/3= EI

2 1

2

1

(4/9)+(4/3)

3.753.756.75

23


45/148

45

=-6.40

6.92

Member 1 : [q]1 = [k]1[d]1 + [qF]1

MBA

MAB

2

3

= 0.779/EIB

A0

+-6.75

6.75

Substitute B and Cin the member matrix,

1

2

wL2/12 = 6.75wL2/12 = 6.75

1 kN/m

9 m 1[FEF]

4.56 kN 4.44 kN

6.92 kNm 6.40 kNm

wL2/12 = 6.75wL2/12 = 6.75

= EI4/9

2/9

3

4/9

2/9

2

1 kN/m

19 m

10 kN


46/148

46

Member 2 : [q]2 = [k]2[d]2 + [qF]2

Substitute B and Cin the member matrix,

MCB

MBC

1

2

= EI4/3

4/3

2/3

2/3

2 1

=0

6.40

C

B

= 2.423/EI

= 0.779/EI

2

12

[FEF]

PL/8 = 3.75PL/8 = 3.75

1.5 m1.5 m

2

7.13 kN 2.87 kN

6.40 kNm

10 kN

2

PL/8 = 3.75PL/8 = 3.75

+-3.75

3.75

1 kN/m

10 kN


47/148

47

A C

B

10 kN1 kN/m

9 m 1.5 m1.5 m

6.92 kNm

11.57 kN 2.87 kN4.56 kN

4.56 kN 4.44 kN 7.13 kN 2.87 kN

1 kN/m

6.92 6.406.40

M

(kNm) x (m)

-6.92 -6.40

3.48 4.32

V(kN)x (m)

4.56

-4.44-2.87

4.56 m

7.13

B = +0.779/EIC= +2.423/EI


48/148

48

20 kN

A CBEI2EI

4 m 4 m

9kN/m 40 kNm

Example 2


(a) Determine the deflection and rotation atB.

(b) Determine all the reactions at supports.

2 4 62 6


49/148

49

1 2 3

0.5625

3

-0.375

-0.375

1 3 5

[K] =EI

3 4

3

4

Use 4x4 stiffness matrix,

[k]1

0.375EI

-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

EI

-0.75EI

2EI

0.75EI

0.75EI

-0.375EI

EI

-0.75EI

-0.75EI

1 2 3 4

1

2

3

4

[k]2

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI- 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

3 4 5 6

3

4

5

6

2 EI EI

1 2

1 5

i j j

Mi

i

Vj

Mj

Vi

[ ]

=

LEILEILEILEI

LEILEILEILEI

LEILEILEILEI

LEILEILEILEI

k

/4/6/2/6

/6/12/6/12

/2/6/4/6

/6/12/6/12

22

2323

22

2323

20 kN20 kN


50/148

50

9kN/m 40 kNm

4 m 4 m

12 kNm

18 kN

12 kNm

18 kN 18 kN

D4

D3

Global:

D4

D3=

9.697/EI

-61.09/EI

Q4 = 40

Q3 = -203

4+

-12

18 3

4

40 kNm

12 kNm

1 2

1

2

3

4

5

6

1 2 3

2EI EI

[Q] = [K][D] + [QF]

0.5625

3

-0.375

-0.375

=EI

3 4

3

4

9 kN/m

2 4


51/148

51

Member 1:

1

9 kN/m

A B

12 kNm

[qF]1

12 kNm

18 kN 18 kN

1

9 kN/m

A B

[q]1

53.21 kNm

48.18 kN

67.51 kNm12.18 kN

1

1 3

1 2

2EI 12 kNm 12 kNm

18 kN 18 kN

q2

q1

q4L

q3L

67.51

48.18

53.21

-12.18

12

18

-12

18

12(2EI)/43

-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

EI-0.75EI

2EI

0.75EI

0.75EI

-0.375EI

EI

-0.75EI

-0.75EI

1 2 3 4

1

2

3

4

[q]1 = [k]1[d]1 + [q

F

]1

d3 = -61.09/EI

d4 = 9.697/EI

d2 = 0

d1 = 0

4 6


52/148

52

2B C

[q]2

Member 2:

[q]2 = [k]2[d]2 + [qF]2

2

3 5

2 3

EI

q4R

q3R

q6

q5

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI- 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

3 4 5 6

3

4

5

6

0

0

0

0

d3 =-61.09/EI

d4 = 9.697/EI

d6 = 0

d5 = 0

18.06 kNm

7.82 kN

13.21 kNm

7.82 kN

-13.21

-7.818

-18.06

7.818

9 kN/m 53.21 kNm

18.06 kNm13.21 kNm


53/148

53

V(kN)x (m)

-7.818

48.18

-7.818-

M

(kNm)x (m)

13.21

-18.08-

-67.51

-

1A B

[q]148.18 kN67.51 kNm

12.18 kN

2B C

[q]27.82 kN7.82 kN

12.18+

53.21

+

20 kN

4 m 4 m

9kN/m 40 kNm

7.818 kN

18.06 kNm67.51 kNm

48.18 kN

D3=

B= -61.09/EI

D4 = B = +9.697/EI


54/148

20 kN

9kN/m

10 kN


55/148

55

Global1 2

1

2

3

4

5

6

1 2 3

A CB EI2EI

9kN/m 40 kNm

4 m 2 m 2 m

10 kN

5 kNm

5 kN

5 kNm

5 kN

2

12 kNm

18 kN

12 kNm

18 kN

9 kN/m

1[FEF]

i j j

Mi

i

Vj

Mj

Vi

[ ]

=

LEILEILEILEI

LEILEILEILEI

LEILEILEILEI

LEILEILEILEI

k

/4/6/2/6

/6/12/6/12

/2/6/4/6

/6/12/6/12

22

2323

22

2323

44

2 4 62


56/148

56

1 2

1 3 5

1 2 3

4 m 2 m 2 m

1 5

1 2

0.375

0.5

0.375 0.5 1

[K] = EI

3 4 6

3

4

6

0.5625

3

-0.375

-0.375

12(2EI)/43

[k]1 =-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

EI

-0.75EI

2EI/L

0.75EI

0.75EI

-0.375EI

EI

-0.75EI

-0.75EI

1 2 3 41

2

3

4

[k]2 =0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

0.1875EI

-0.375EIEI0.375EI - 0.1875EI

0.1875EI

0.375EI0.5EI

-0.375EI

EI

3 4 5 6

3

4

5

6

20 kN

9kN/m 40 kN

10 kN20 kN

40 kN


57/148

57

MBA+MBC= 40

VBL+VBR = -20

MCB = 0B

B

C-12 + 5 = -7-5

B

B

C

= -7.667/EI-116.593/EI

52.556/EI

Global: [Q] = [K][D] + [QF]

= EI

0.5625

3

-0.375

-0.375

0.375

0.50.375 0.5 1

3 4 6

3

46

A CB

9kN/m 40 kNm

21

1

2

3

4

5

6

1 2 310 kN5 kNm

5 kN

5 kNm

5 kN

2

12 kNm

18 kN

12 kNm

18 kN

9 kN/m

1[FEF]

40 kNm

5 kNm5 kNm

5 kN

12 kNm

18 kN

+18 + 5 = 23

1

2

3

4

12 kNm9 kN/m

12 kNm


58/148

58

0.375EI

=

-0.75EI2EI

0.75EI - 0.375EI

0.375EI

0.75EI

EI

-0.75EI

2EI/L

0.75EI

0.75EI

-0.375EI

EI

-0.75EI

-0.75EI

1 2 3 4

1

2

3

4

=

91.78

55.97

60.11

-19.97

91.78 kNm

55.97 kN

Member 1: [q]1 = [k]1[d]1 + [qF]1

MAB

VA

MBA

VBL +

12

18

-12

18=-116.593/EI

=-7.667/EI

0

0

B

B

1

1 3

1 2

12 kNm

18 kN

12 kNm

18 kN1

[FEF]

19.97 kN

60.11 kNm4 m

9kN/m

1

12 kNm

18 kN

12 kNm

18 kN18 kN

10 kN5 kNm

k

64

5 kNmk


59/148

59

VC

MBC

VBR

MCB

+

5

5

5

-5

=

-20.11

-0.0278

0

10.03

=52.556/EI

=-116.593/EI

=-7.667/EIB

B

C

0

5 kN m

5 kN

5 kNm

5 kN

2

[FEM]

10.03 kN

2 m

10 kN

2 m

2

=

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

3 4 5 6

3

4

5

6

0.0278 kN

20.11 kNm

Member 2: [q]1 = [k]1[d]1 + [qF]1

223

535 kN m

5 kN

5 kNm

5 kN

91.78 kNm9kN/m 20.11 kNm

2 m

10 kN

2 m91.78 kNm20.11 kNm


60/148

60

20 kN

4 m 2 m

9kN/m 40 kNm

10 kN

2 m

10.03 kN

91.78 kNm

55.97 kN

V(kN)x (m)

55.97

-0.0278

19.97

+

-10.03 -10.03

-

55.97 kN 19.97 kN60.11 kNm4 m

1

10.03 kN0.0278 kN

2

60.11 kNm

M

(kNm) x (m)

-91.78

-

+20.11

60.11

B = -7.667/EI

C= 52.556/EI

B = -116.593/EI

Example 4


61/148

61

40 kN

8 m 4 m 4 m

2EI EI

6 kN/m

AB

C

B = -10 mm

p

For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.

(b) Draw the quantitative free-body diagram of member.

(c) Draw the quantitative shear diagram, bending moment diagram

and qualitative deflected shape.

TakeI= 200(106) mm4 andE= 200 GPa and supportB settlement 10 mm.

1 2 3

2 EI EI

1


62/148

62

[k]1 =4

8

8

4

1 2

1

2

EI8

1 2

2 EI EI

2

42

2 3

2

3

[K] = EI812

2

4

4

2

2 3

2

3

[k]2 =EI8

Use 2x2 stiffness matrix:

i jMiMj[ ]

= LEILEI

LEILEI

k /4/2

/2/422

40 kN6 kN/m

B C

1 2 3


63/148

63

75 kNm

6(2EI)/L2 = 75 kNm

10 mm

137.5 kNm

6(EI)/L2

= 37.5 kNm

10 mm

2

[FEM]

6(EI)/L2

= 37.5 kNm

D2

D3

=

-61.27/EI rad

185.64/EI rad

2EI EIA B = -10 mm 1 2

[FEM]load

32 kNmwL2/12 = 32 kNm

6 kN/m

1 40 kNmPL/8 = 40 kNm2

40 kN

8 m 4 m 4 m

32 kNm 40 kNmPL/8 = 40 kNm

75 kNm37.5 kNm

+

-32 + 40 + 75 -37.5 = 45.5Q2= 0

Q3 = 0 -40 - 37.5 = -77.5

D2

D3

2

42

2 3

2

3

12

8

EI

=


[FEM]load

6 kN/m2

2EI

1


64/148

64

6(2EI)/L2 = 75 kNm

32 kNmwL2/12 = 32 kNm1

75 kNm

6(2EI)/L2 = 75 kNm10 mm

1[FEM]

6 kN/m

18 m

1

q1

q2

d1 = 0

d2 = -61.27/EI=

76.37 kNm

-18.27 kNm

Member 1: [q]1 = [k]1[d]1 + [qF]1

32 kNmwL2/12 = 32 kNm

75 kNm

+32 + 75 = 107

-32 + 75 = 43

31.26 kN

76.37 kNm

16.74 kN

18.27 kNm

4

8

8

4

1 2

1

28

EI=

2 3 40 kN

[FEM]


65/148

65

Member 2: [q]2 = [k]2[d]2 + [qF]2

2 40 kNmPL/8 = 40 kNm 2

37.5 kNm

6(EI)D/L2 = 37.5 kNm

10 mm

2

[FEM]load

[FEM]

40 kNm

37.5 kNm

6(EI)D/L2 = 37.5 kNm

PL/8 = 40 kNm

=18.27 kNm

0 kNm

q2

q3

+40 - 37.5 = 2.5

-40 - 37.5 = -77.5

d2 = -61.27/EI

d3 = 185.64/EI

17.72 kN

18.27 kNm

22.28 kN

40 kN

2

2

4

4

2

2 3

2

38

EI=

2 EI EI

264


66/148

66

Alternate method: Use 4x4 stiffness matrix

1 253

1

8

EI= 1.5

1.5

-0.375

12(2)/82

4

-1.5

8

1.5

-1.5

-1.5

-0.375

0.375

1.5

4

-1.5

8

1 2 3 4

1

2

3

4

[k]1

8

EI=

0.75

0.75

-0.1875

12/82

2

-0.75

4

0.75

-0.75

-0.75

-0.1875

0.1875

0.75

2

-0.75

4

3 4 5 6

3

4

5

6

[k]2

2

-0.75

-0.1875

-0.75

0.1875

-0.75

0.75

2

-0.75

4

-0.1875

0.75

1.5 4

1.5

4

0.375

-0.3751.5

1.5

8-1.5

-0.375

-1.5

0

0

0

0

0

0

0

0

-0.75

12

0.5625

-0.758

EI=[K]

1

23

4

5

6

1 4 5 62 3

40 kN6 kN/m

B C

264

2


67/148

67


8 m 4 m 4 m

2EI EIA B = -10 mm 12

531

40 kN40 kNm

20 kN

40 kNm

20 kN

232 kNm

24 kN

32 kNm

24 kN

6 kN/m

1[FEF]

531

D4

D6 =

-61.27/EI= -1.532x10-3 rad

185.64/EI= 4.641x10-3 rad

+(200x200/8)(-0.75)(-0.01) = 37.5

(200x200/8)(0.75)(-0.01) = -37.5

D4

D6

Q4= 0

Q6= 0

2

4

12

2+

8

-40

4 6

4

68

EI=

-0.75

-0.75 D5 = -0.01)8

200200(

+ 4

6

5

D4

D6

Q4= 0

Q6= 0

2

4

12

2

4 64

68

EI= +

8

-40

24

32 kNm6 kN/m


68/148

68

+

32

24

-32

24

q1

q2

q4

q3

1.5

1.5

-0.375

12(2)/82

4

-1.5

8

1.5

-1.5

-1.5

-0.375

0.375

1.5

4

-1.5

8

1 2 3 4

1

2

34

=(200x200)

8

q1

q2

q4

q3=

76.37 kNm

31.26 kN

-18.27 kNm

16.74 kN

Member 1: [q]1 = [k]1[d]1 + [qF]1

13

1 B = -10 mm32 kNm

24 kN

32 kN m

24 kN

1[FEF]load

d2 = 0

d1 = 0

d4 = -1.532x10-3

d3 = -0.01

6 kN/m

18 m

31.26 kN

76.37 kNm

16.74 kN

18.27 kNm

6440 kN

40 kNm40 kNm


69/148

69

Member 2:

+40

20

-40

20

q3

q4

q6

q5

d4 = -1.532x10-3

d3 = -0.01

d6 = 4.641x10-3

d5

= 0

0.75

0.75

-0.1875

12/82

2

-0.75

4

0.75

-0.75

-0.75

-0.1875

0.1875

0.75

2

-0.75

4

3 4 5 63

4

5

6

=(200x200)

8

q3

q4

q6

q5=

18.27 kNm22.28 kN

0 kNm

17.72 kN

17.72 kN

18.27 kNm

22.28 kN

40 kN

2

2

53

-10 mm = B

20 kN20 kN

2

[FEF]

40 kN6 kN/m

B C76.37 kNm


70/148

70

8 m 4 m 4 m

2EI EIA B = -10 mm31.26 kN 16.74 + 22.28 kN 17.72 kN

V(kN)x (m)

+ +--

31.26

-16.74

22.28

-17.725.21 m

M

(kNm) x (m)

-76.37

5.06

-18.27

70.85

+

--

D6 = C= 4.641x10-3 rad

d4 = B = -1.532x10-3

rad

Deflected

Curve

B = -10 mm

Example 5


71/148

71

For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.


(c) Draw the quantitative shear diagram, bending moment diagram


TakeI= 200(106) mm4 andE= 200 GPa and support Csettlement 10 mm.

4 kN

8 m 4 m 4 m

2EI EI

0.6 kN/m

AB C

C= -10 mm

20 kNm

2 EI EI

264

2


72/148

72

12

-10 mm53

15

1

Q3

Q6

Q4

0.75 2

0.75

2

4

1.5

1.5

-0.375

12(2)/82

4

-1.5

8

1.5

-1.5

-1.5

-0.375

0.375

1.5

4

-1.5

8

1 2 3 4

1

2

3

4

[k]1

8

EI=

8

EI=

0.75

0.75

-0.1875

12/82

2

-0.75

4

0.75

-0.75

-0.75

-0.1875

0.1875

0.75

2

-0.75

4

3 4 5 6

3

4

5

6

[k]2

D3

D6

D4

-0.1875

-0.75

-0.75 D5 = -0.01


Member stiffness matrix [k]4x4

0.5625

-0.75

-0.75

12

3

4

6

3 4 6

8

EI= )

8

200200(

+

3

4

6

5

QF3

QF6

QF4+

4 kN0.6 kN/m

A

BC

20 kNm2 EI EI

1 2

642


73/148

73

4 kN4 kNm

2 kN

4 kNm

2 kN

23.2 kNm

2.4 kN

3.2 kNm

2.4 kN

0.6 kN/m

1[FEF]

8 m 4 m 4 m2EI

EIA10 mm

1 2

3 51

3

4

6

0.5625

0.75

-0.75

-0.75

12

2

0.75

2

3 4 6

48

EI=

D3

D6

D4 +

9.375

37.5

37.5

2.4+2 = 4.4

-4.0

-3.2+4 = 0.8+

D3

D6

D4

-377.30/EI= -9.433x10-3 m

+74.50/EI= +1.863x10-3 rad

-61.53/EI= -1.538x10-3 rad=


Q3 = 0

Q6 = 20

Q4 = 0

24

3 2 kNm3.2 kNm

0.6 kN/m


74/148

74

+3.2

2.4

-3.2

2.4

q1

q2

q4

q3

q1

q2

q4

q3=

43.19 kNm

8.55 kN

6.03 kNm

-3.75 kN

Member 1: [q]1 = [k]1[d]1 + [qF]1

d2 = 0

d1 = 0

d4 = -1.538x10-3

d3 = -9.433x10-3

0.6 kN/m

18 m

8.55 kN

43.19 kNm

3.75 kN

6.03 kNm

13

1

3.2 kNm

2.4 kN2.4 kN

1[FEF]load 8 m

1.5

1.5

-0.375

12(2)/82

4

-1.5

8

1.5

-1.5

-1.5

-0.375

0.375

1.5

4

-1.5

8

1 2 3 4

1

2

3

48

200200=

644 kN

4 kNm4 kNm

[FEF]


75/148

75

Member 2:

+4

2

-4

2

q3

q4

q6

q5

d4 = -1.538x10-3

d3 = -9.433x10-3

d6 = 1.863x10-3

d5 = -0.01

q3

q4

q6

q5=

-6.0 kNm

3.75 kN

20.0 kNm

0.25 kN

0.25 kN

6.0 kNm

3.75 kN

4 kN

2

10 mm253 2 kN2 kN

2

[FEF]

8 m

0.75

0.75

-0.1875

12/82

2

-0.75

4

0.75

-0.75

-0.75

-0.1875

0.1875

0.75

2

-0.75

4

3 45

63

4

5

6

8

200200=

20 kNm

4 kN0.6 kN/m

BC

20 kNm


76/148

76

8 m 4 m 4 m

2EI EIA

C

10 mm

Deflected

Curve

0.6 kN/m

18 m

8.55 kN

43.19 kNm

3.75 kN

6.03 kNm

V(kN)

x (m)+

8.55

3.75

-0.25

0.25 kN

6.0 kNm

3.75 kN

4 kN

2

20 kNm

M

(kNm) x (m)

-43.19

621

+

-

20

D3 = -9.433 mm

D6

= +1.863x10-3 radD4

= -1.538x10-3 rad

D5 = -10 mm

Example 6

Internal Hinges


77/148

77

30 kN

A CB

EI2EI

4 m 2 m

9kN/m

Hinge

2 m

For the beam shown, use the stiffness method to:(a) Determine all the reactions at supports.

(b) Draw the quantitative shear and bending moment diagrams


E= 200 GPa,I= 50x10-6 m4.

C2EI EI

43

2

3 4


78/148

78

[K] = EI1

2

1 2

0.0

0.0

1.0

2.0

AC

B4 m 2 m 2 m

1 21

[k]1 =EI

2EI

2EI

EI

3 1

3

1

[k]2 =0.5EI

1EI

1EI

0.5EI

2 4

2

4

Use 2x2 stiffness matrix,

i j

MiMj

[ ]

=LEILEI

LEILEIk

/4/2

/2/422

30 kN

9kN/m

Hinge


79/148

79

0.0

0.0

D1

D2

15 kNmB C

15 kNm30 kN

2

12 kNm

A B

12 kNm 9kN/m

1[FEF]

= EI1

2

1 2

0.0

0.0

1.0

2.0

D1

D2

=

0.0006 rad

-0.0015 rad

-12

15

+

Global matrix:

A C

B

1 2

43

12 kNm 15 kNm1

2

12 kNm12 kNm 9kN/m3

1


80/148

80

Member 1:

q3

q1

12

-12+=

EI

2EI

2EI

EI

3 1

3

1

18

0.0=

d3

d1

= 0.0

= 0.0006

A B1[FEF]A

B

1

18 kNm22.5 kN 13.5 kN

A B

9 kN/m

1

CB

415 kNm

B C15 kNm 30 kN


81/148

81

B C

30 kN

22 2

q2

q4

= -0.0015

= 0.0

d2

d4

15

-15+=

0.5EI

1EI

1EI

0.5EI

2 4

2

4

0.0

-22.5

=

Member 2:

22.5 kNm

20.63 kN9.37 kN

22.5 kNmB C

30 kN

A B

9 kN/m

9.37 kN13.5 kN


82/148

82

30 kN

A CB

EI2EI

4 m 2 m

9kN/m

Hinge

2 m

-20.63

V(kN)x (m)

22.59.37

-13.5

+

2.5 m

M

(kNm)x (m)

-18

10.13

-22.5

18.75

+

-

+-

-

BL= 0.0006 radBR= -0.0015 rad

13.5 kN22.5 kN 20.63 kN9.37 kN18 kNm 22.87 kN

Example 7


83/148

83

20 kN

A CBEI2EI

4 m 4 m

9kN/m

Hinge




E= 200 GPa,I= 50x10-6 m4.

2

2 EI EI 6

7

4

5

4

5

6

7

1 3


84/148

84

A CB1

0.375

0.375

1.0

-0.75

-0.75 2.0

0.0

0[K] = EI

1

2

3

1 2 3

0.5625

[k]1 =

0.75EI

EI

-0.75EI

2EI/L

0.375EI

0.75EI

0.75EI

-0.375EI

2EI

0.75EI

EI

-0.75EI

-0.75EI

- 0.375EI

0.375EI

-0.75EI

4 5 1 2

4

5

1

2

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

[k]2 =

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

1 3 6 7

1

3

6

7

2

20 kN

A C

B9kN/m

5

20 kN


85/148

85

Q1 = -20

Q3 = 0.0

Q2 = 0.0

D1

D3

D2

18

0.0

-12+

D1

D3

D2 =

-0.02382 m

0.008933 rad

-0.008333 rad

[FEM]

12 kNmA B

12 kNm

18 kN18 kN

1

9kN/mA CB

1 2

6

7

4 2 EI EI

Global:

= EI

1

2

3

1

0.5625

0.375

-0.75

2

-0.75

2.0

0.0

3

0.375

0

1.0

12 kNm

18 kN

1

2

3

2EI1

24

5

12 kNmA B

12 kNm

1

9kN/m


86/148

86

A B1 [FEF] 18 kN18 kN

+12

18

-12

18

Member 1:

q4

q5

q1

q2

=

0.75EI

EI

-0.75EI

2EI/L

0.375EI

0.75EI

0.75EI

-0.375EI

2EI

0.75EI

EI

-0.75EI

-0.75EI

- 0.375EI

0.375EI

-0.75EI

4 5 1 2

4

5

1

2

= 0.0

= 0.0

= -0.02382

= -0.00833

d5

d4

d2

d1

=107.32

44.83

0.0

-8.83

A B

107.32 kNm

8.83 kN

44.83 kN

9kN/m

1

CB

2

EI1

3

6

7


87/148

87

Member 2:

CB 3

44.66 kNm

B C

11.16 kN11.16 kN

2

+0

0

0

0

= -0.02382

= 0.008933

= 0.0

= 0.0

d3

d1

d7

d6

q1

q3

q7

q6

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

=

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

1 3 6 7

1

3

6

7

=0.0

-11.16

-44.66

11.16

20 kN

9kN/m


88/148

88

A CB

4 m 4 m

V(kN)x (m)

44.83

8.83

-11.16-11.16

+

-

M(kNm)

x (m)

-107.32

-44.66-

-

A B107.32 kNm

8.83 kN44.83 kN

9kN/m

144.66 kNm

B C

11.16 kN11.16 kN

2

BL= -0.008333 rad BR= 0.008933 rad

Example 8

F th b h th tiff th d t


89/148

89

30 kN

A C

BEI

2EI

4 m 2 m

9kN/m

Hinge

40 kNm

2 m




40 kNm at the end of memberAB. E= 200 GPa,I= 50x10-6 m4.

A C2EI EI

1 2

433 4

1

2


90/148

90

[K] = EI1

2

1 2

0.0

0.0

1.0

2.0

B4 m 2 m 2 m

1 2

[k]1 =EI

2EI

2EI

EI

3 1

3

1

[k]2 =0.5EI

1EI

1EI

0.5EI

2 4

2

4

1

Use 2x2 stiffness matrix: i jMiMj

[ ]

=

LEILEI

LEILEIk

/4/2

/2/422

3

30 kN

A C

EI2EI

9kN/m

Hinge

40 kNm40 kNm


91/148

91

15 kNmB C

15 kNm30 kN

2

12 kNm

A B

12 kNm 9kN/m

1[FEM]

Global matrix:

A C

B

1 2

43

1

2

12 kNm 15 kNm

BEI2EI 40 kNm

Q1 = 40

Q2 = 0.0

D1

D2

-12

15

+

D1

D2=

0.0026 rad

-0.0015 rad

= EI1

2

1 2

0.0

0.0

1.0

2.0

40 kNm

12 kNm

A B

12 kNm 9kN/m

1[FEF]A 1

3

1


92/148

92

A B1[FEF]AB

1

40 kNmA B38 kNm

9 kN/m

1.5 kN37.5 kN

Member 1:

38

40

=q3

q1

= 0.0

= 0.0026

d3

d1

12

-12

+=EI

2EI

2EI

EI

3 1

3

1

CB

2

4

2

15 kNmB C

15 kNm 30 kN

2


93/148

93

2 2

Member 2:

q2

q4

= -0.0015

= 0.0

d2

d4

15

-15

+=0.5EI

1EI

1EI

0.5EI

2 4

2

4

0.0

-22.5=

22.5 kNmB C

30 kN

20.63 kN9.37 kN

9.37 kN

1.5 kN

7 87 kN

40 kNmA B

38 kNm

9 kN/m

1 5 kN37 5 kN

22.5 kNmB C

30 kN

20 63 kN9 37 kN


94/148

94

30 kN

A CB

4 m 2 m

9kN/m

Hinge

40 kNm

2 m

-20.63

V(kN)x (m)

37.5

9.371.5+

M

(kNm)x (m)

-38

40

-22.5

+

-

18.75

+--

BL= 0.0026 radBR=- 0.0015 rad

7.87 kN38 kN m 1.5 kN37.5 kN 20.63 kN9.37 kN

Example 9



95/148

95

20 kN

A CB

EI2EI

4 m 4 m

9kN/m

Hinge40 kNm




40 kNm at the end of memberAB. E= 200 GPa,I= 50x10-6 m4

20 kN

EI2EI

9kN/m

Hinge40 kNm


96/148

96

A C

B1 2

6

7

4

5

1

2

3

A CBEI2EI

4 m 4 m

Hinge40 kN m

12 kNm

18 kN

12 kNm

18 kN

9 kN/m

1[FEM]

i j j

Mi

i

Vj

Mj

Vi

[ ]

=

LEILEILEILEI

LEILEILEILEI

LEILEILEILEI

LEILEILEILEI

k

/4/6/2/6

/6/12/6/12

/2/6/4/6

/6/12/6/12

22

2323

22

2323

44

1 22A C

B

6

7

4

5

6

7

4

5

1

2 EI EI 1

2

3


97/148

97

[K] = EI

1

2

3

1 2 3

0.5625

0.0

0

0.375

0.375

1.0

-0.75

-0.75

2.0

[k]1 =

0.75EI

EI

-0.75EI

2EI

0.375EI

0.75EI

0.75EI

-0.375EI

2EI

0.75EI

EI

-0.75EI

-0.75EI

- 0.375EI

0.375EI

-0.75EI

4 5 1 2

4

5

1

2

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

[k]2 =0.1875EI

-0.375EIEI0.375EI - 0.1875EI

0.1875EI

0.375EI0.5EI

-0.375EI

EI

1 3 6 7

1

3

6

7

75

20 kN

EI

9kN/m

40 kNm

20 kN

40 kNm

1


98/148

98

Q1 = -20

Q3

= 0.0

Q2 = 40

D1

D3

D2

18

0.0

-12+

D1

D3

D2 =

-0.01316 m

0.0049333 rad

-0.002333 rad

A C

B1 2

64 2 EI EI

12 kNm

18 kN

12 kNm

18 kN

9 kN/m

1[FEF]

= EI

1

2

3

1 2 30.5625

0.0

0

0.375

0.375

1.0

-0.75

-0.75

2.0

12 kNm

18 kN

Global:

1

2

3

1

12

1

2EI4

5

12 kNm

18 kN

12 kNm

18 kN

9 kN/m

1[FEF]12 kNm

18 kN

12 kNm

18 kN

1


99/148

99

1

2EI

[q]1

+12

18

-12

18

q4

q5

q1

q2

=

0.75EI

EI

-0.75EI

2EI

0.375EI

0.75EI

0.75EI

-0.375EI

2EI

0.75EI

EI

-0.75EI

-0.75EI

- 0.375EI

0.375EI

-0.75EI

4 5 1 2

4

5

1

2

=87.37

49.85

40

-13.85

Member 1: [q]1 = [k]1[d]1 + [qF]1

13.85 kN

40 kNm87.37 kNm

49.85 kN

= 0.0

= 0.0

= -0.01316= -0.002333

d5

d4

d2

d1

22 C

B

1

3

EI 6

7


100/148

100

Member 2: [q]2 = [k]2[d]2 + [qF]2

+0

0

0

0

q1

q3

q7

q6

0.375EI

0.375EI

-0.1875EI

0.5EI

-0.375EI

-0.375EI

=

0.1875EI

-0.375EIEI

0.375EI - 0.1875EI

0.1875EI

0.375EI

0.5EI

-0.375EI

EI

1 3 6 71

3

6

7

=0.0

-6.18

-24.696.18

24.69 kNm

6.18 kN6.18 kN

= -0.01316

= 0.004933

= 0.0

= 0.0

d3

d1

d7

d6

22

[q]2

20 kN

A CEI2EI

9kN/m 40 kNm

BL= -0.002333 rad BR = 0.004933 rad


101/148

101

A CB4 m 4 m

24.69 kNmB C

6.18 kN6.18 kN

A B

87.37 kNm 13.85 kN49.85 kN

40 kNm

M(kNm)

x (m)

-

-87.37

+40

--24.69

BL 0.002333 rad BR 0.004933 rad

49.85

-6.18

V(kN)x (m)

-6.18-

13.85+

Fixed-End Forces (FEF)

- Axial- Bending

Temperature Effects


102/148

102

Bending

Curvature

Tm> TR

Thermal Fixed-End Forces (FEF)

Tt

bt TTT+=

Tt F

AFF

BF

)(tand

T

d

TT tb =

=

Room temp = TR

ctd

TR

NA


103/148

103

Tl

2mT =

Tb > Tt BF

BMF

AM

Rmaxial TTT = )(

tbbending TTT = )(

)()(d

TyTy

=

y

A

A B

axialaxial

A B

y

y

TR Tm

Tt

Tb

Tb - Tt

cbd

- Axial

axialaxial

F

TtT > T

TR Tm


104/148

104

A B

F

AFF

BF

dAFA

axial

F

A =

dAE axial=

dATE axial = )(

axialTEA )(=

AETTF RmAF

axial )()( =

dATE axial= )(

Tb > Tt

Tm> TR

Rmaxial TTT = )(

- Bending

y

y

)()(

d

TyTy

=

y Tt


105/148

105

A B

y

dAyMA

y

F

A =

dATyE y = )(

dAd

TyyE

= )(

= dAydTE 2)(

EId

TTF ulA

F

bending )()(

=

F

BMF

AM

dAyE=

tbbending TTT = )(

y

Tb

O

d

)()( dd

Elastic Curve: Bending


106/148

106

ds

dx

After

deformation

d

Before

deformation

y dx

ds = dx y

)()( ddx =

)( dy )(1

dx

d

=

Bending Temperature

dx

Tt

Tb

Tl > Tu

O


107/148

107

dxd

Tyyd )()(

=

dxd

Td )()(

=

Tb

Tt

yy

dT

y

Tl > Tu

Tu

d

2

bt

m

TT

T

+

= d

MM

Tb

cb

ct

Tt

T= Tb- Tt

d

T=

EI

M

d

T

dx

d=

== )(

1)(

Example 10



108/148

108

A CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm



Room temp = 32.5oC, = 12x10-6 /oC,E= 200 GPa,I= 50x10-6 m4.

A CB

EI2EI

T2 = 40oC

T1 = 25oC

182 mm


109/148

109

12 3

1 2

Mean temperature = (40+25)/2 = 32.5

Room temp = 32.5oC

FEM+7.5 oC

-7.5 oC

a(T /d)EI19.78 kNm = 19.78 kNm

4 m 4 m

mkNEId

TFFbending =

=

= 78.19)502002)(

182.0

2540)(1012()2)(( 6

12 3

A C

BEI2EI1 2

19.78 kNm19.78 kNm

182 mm


110/148

110

[M1] = 3EI1 + (1.978x10-3)EI

0

Element 1:

M1

M2

q1

q2

[q] = [k][d] + [qF]

+1.978

-1.978(10-3)EI=

1

2

2

1

2 1

2

1EI

Element 2:

M3

M1

q3

q1

+0

0

=0.5

1

1

0.5

1 3

1

3

EI

1= -0.659x10

-3 rad

4 m 4 m

12 3

A C

B

EI2EI1 2

19.78 kNm19.78 kNm

182 mm


111/148

111

Element 2:

M3

M1

q3

q1

= +0

0

0.5

1

1

0.5

1 3

1

3EI

= -0.659x10-3

= 0

=6.59 kNm

-26.37 kNm

= -0.659x10-3

= 0=

-3.30 kNm

-6.59 kNm

Element 1:

M1

M2=

q1

q2+

19.78

-19.78

1

2

2

1

2 1

2

1EI

4.95 kN4.95 kN

26.37 kNm 6.59 kNm

A B

6.59 kNm 3.3 kNm

B C

2.47 kN2.47 kN

4 m 4 m

A CB

4 m 4 m

+7.5 oC

-7.5o

C 3.3 kNm

26.37 kNm

4.95 kN 2.47 kN 2.47 kN


112/148

112

4 m 4 m

M

(kNm)x (m)

26.37

6.59

-3.30

+

-

V(kN) x (m)

-4.95-2.47

- -

B= -0.659x10-3 radDeflected

curve x (m)

Example 11




113/148

113

A CB

EI,AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm



Room temp = 28 oC, a = 12x10-6 /oC,E= 200 GPa,I= 50x10-6 m4,

A = 20(10-3

) m2

A

CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC 1

2 3

7

8

9

1 2

4

5

6


114/148

114

4 m 4 m

Element 1:

mkNm

mkNm

L

AE/)10(2

)4(

)/10200)(1020(2)2( 62623

==

mkN

m

mmkN

L

EI=

=

)10(20

)4(

)1050)(/10200(24)2(4 34626

mkNm

mmkN

L

EI=

=

)10(10)4(

)1050)(/10200(22)2(2 34626

kNm

mmkN

L

EI)10(5.7

)4(

)1050)(/10200(26)2(63

2

4626

2 =

=

mkNm

mmkN

L

EI/)10(75.3

)4(

)1050)(/10200(212)2(12 33

4626

3=

=

A CB

EI,AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm


115/148

115

Fixed-end forces due to temperatures

Mean temperature(Tm) = (40+25)/2 = 32.5oC,

TR = 28 oC

4 m 4 m

mkNEId

TFFbending =

=

= 78.19)502002)(

182.0

2540)(1012()2)(( 6

19.78 kNm19.78 kNm

432 kN432 kN

A B

+12 oC

-3 oC

kNmkNmAETFFaxial 432)/10200)(310202)(285.32)(1012()(2626 ===

A CB

EI,AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm

1

2 3

7

8

9

1 2

4

5

6


116/148

116

FEM

+12 oC

-3 oC

19.78 kNm

A B

19.78 kNm

432 kN432 kN

Element 1:

[q] = [k][d] + [qF]

0

0

0

d1

d2

d3

q4

q6

q5

q1

q2q3

=

4

5

6

1

2

3

4 1

- 2x106

0.00

0.00

2x106

0.00

0.00

2

0.00

-3750

-7500

0.00

3750

-7500

3

0.00

7500

10x103

0.00

-7500

20x103

2x106

0.00

0.00

-2x106

0.00

0.00

5

0.00

3750

7500

0.00

-3750

7500

6

0.00

7500

20x103

0.00

-7500

10x103

432

-19.78

0.00

-432

0.00

19.78

+

4 m 4 m

A CB

EI,AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm

1

2 3

7

8

9

1 2

4

5

6


117/148

117

Element 2:

[q] = [k][d] + [qF]

d1

d3

d2

0

0

0

0

0

0

0

0

0

+

q1

q3

q2

q7

q8q9

- 1x106

0.00

0.00

1x106

0.00

0.00

0.00

-1875

-3750

0.00

1875

-3750

0.00

3750

5x103

0.00

-3750

10x103

1x106

0.00

0.00

-1x106

0.00

0.00

0.00

1875

3750

0.00

-1875

3750

0.00

3750

10x103

0.00

-3750

5x103

7 8 91 2 3

=

1

2

3

7

8

9

A CB

EI,AE2EI, 2AE

4 m 4 m

T2 = 40oC

T1 = 25oC

182 mm

1

2 3

7

8

9

1 2

4

5

6


118/148

118

D1

D3

D2 =

0.000144 m

-719.3x10-6

rad

-0.0004795 m

Q1 = 0.0

Q3 = 0.0

Q2 = 0.0

D1

D3

D2

-432

19.78

0.0+=

1

2

3

1

3x106

0.0

0.0

2

0.0

5625

-3750

3

0.0

-3750

30x103

Global:

Element 1:

0

0

0

q4

q5

4

5

6

4 1

- 2x106

0.00

0 00

2

0.00

-3750

7500

3

0.00

7500

10 103

2x106

0 00

0.00

5

0.00

3750

7500

6

0.00

7500

20 103

432

19 78

0.00


119/148

119

= 144x10-6

= -479.5x10-6

= -719.3x10-6

0

d1

d2

d3

q6

q1

q2q3

=6

1

2

3

0.00

2x106

0.00

0.00

-7500

0.00

3750

-7500

10x103

0.00

-7500

20x103

0.00

-2x106

0.00

0.00

7500

0.00

-3750

7500

20x103

0.00

-7500

10x103

-19.78

-432

0.00

19.78

+

=

q4

q6

q5

q1

q2q3

144.0 kN

-23.38 kNm

-3.60 kN

-144.0 kN

3.60 kN

9.00 kNm

A B 1

2 3

4

5

6

A B144 kN

3.60 kN

23.38 kNm

144 kN

3.60 kN

9 kNm

= 144x10-6

= -479.5x10-6

719 3 10 6

Element 2:

d1

d

d2

0

0

0

q1

q2

- 1x106

0.00

0 00

0.00

-1875

3750

0.00

3750

5 103

1x106

0 00

0.00

0.00

1875

37500

0.00

3750

10 103

7 8 91 2 3

1

2

3


120/148

120

= -719.3x10-6

144 kN

-9 kNm

-3.6 kN

-144 kN

3.6 kN

-5.39 kNm

=

q1

q3

q2

q7

q8q9

B C 7

8 9

1

2

3

144 kNB C

3.60 kN

9 kNm

144 kN

3.60 kN

5.39 kNm

d3

0

0

0

0

0

0

0

+q3

q7

q8q9

=0.00

1x106

0.00

0.00

-3750

0.00

1875

-3750

5x103

0.00

-3750

10x103

0.00

-1x106

0.00

0.00

37500

0.00

-1875

3750

10x103

0.00

-3750

5x103

3

7

8

9

A CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC

Isolate axial part from the system

RCRA


121/148

121

RA RA=RC

+

Compatibility equation: dC/A = 0

RA = 144 kN

RC = -144 kN

0)4)(285.32(1012)4(

2

)4( 6 =++ E

R

E

R AA

A CB

EI2EI

4 m 4 m

T2 = 40oC

T1 = 25oC


122/148

122

A B

144 kN

3.60 kN

23.38 kNm

144 kN

3.60 kN

9 kNm

144 kN

B C3.60 kN

9 kNm

144 kN

3.60 kN

5.39 kNm

V(kN)x (m)

-3.6

-

M

(kNm) x (m)

23.38

8.98

-5.39

+

-

D3 = -719.3x10-6 radDeflected

curve x (m)

D2= -0.0004795 mm

Skew Roller Support

- Force Transformation

- Displacement Transformation- Stiffness Matrix


123/148

123

j

x

y

Displacement and Force Transformation Matrices

4

5

6


124/148

124

m

i

j

m

i

j

x

y

1

2

3

4

5

6

yx

1

2

3

x y

j

y

x

45

6

y

x

Force Transformation

yx qqq coscos 5'4'4 =

xy qqq coscos 5'4'5 =

6'6 qq =


125/148

125

i

yx

1

2

3

m

i

j

x

y

1

2

3

4

5

6

xx ijx

=yy ij

y

=

66 qq

6

5

4

q

q

q

=

100

0

0

xy

yx

6'

5'

4'

q

q

q

=

6'

5'

4'

3'

2'

1'

6

5

4

3

2

1

1000000000

0000

000100

00000000

qq

q

q

qq

qq

q

q

qq

xy

yx

xy

yx

[ ] [ ] [ ]'qTq T=

y

d4

d4cos

x

x

x

Displacement Transformation

4

5

6j4

5

6x y

yx ddd coscos' 544 +=

xy ddd' coscos 545 +=

66' dd =


126/148

126

x

y

yx

x

y

xd4c

osy

y

d5

x

y

d5cos

x

d5co

sy

j66

6'

5'

4'

d

d

d

=

100

0

0

xy

yx

6

5

4

d

d

d

=

6

5

4

3

2

1

6'

5'

4'

3'

2'

1'

1000000000

0000

000100

00000000

dd

d

d

dd

dd

d

d

dd

xy

yx

xy

yx

[ ] [ ][ ]dTd ='

[ ] [ ] [ ]'qTq T=

[ ] [ ][ ][ ]

)'( FT

qd'k'T +=

[ ] [ ][ ] [ ] [ ]FTT qTd'k'T '+=


127/148

127

[ ] [ ] [ ][ ][ ] [ ] [ ] [ ][ ] [ ]FFTT qdkqTdTk'Tq +=+= '

[ ] [ ] [ ][ ]TkTkTherefore T ', =

[ ] [ ] [ ]FTF qTq '=

[ ] [ ] [ ]'qTq T=

[ ] [ ][ ]dTd ='

[ ] [ ] [ ][ ]TkTk T '=

i j

ji 4*

5*6*

1*

2*3* 1

Stiffness matrix

i j

4

56

1

2

3


128/148

128

ix = cos i

iy= sin i

jx= cos j

jy= sin j

[q*] = [T]T[q]

*6

5*

*4

*3

2*

*1

q

qq

q

q

q

=

100000

00000000

000100

0000

0000

jxjy

jyjx

ixiy

iyix

1 4 5 62 3

1*

2*

3*

4*5*

6*

[ T]T

6'

5'

4'

3'

2'

1'

q

qq

q

q

q

[ ]

=000100

0000

0000

ixiy

iyix

T

1 4 5 62 3

1

2

3


129/148

129

[ ]

=

100000

0000

0000

jxjy

jyjx

T

4

5

6

[ ]

=

LEILEILEILEI

LEILEILEILEI

LAELAE

LEILEILEILEI

LEILEILEILEI

LAELAE

k

/4/60/2/60

/6/120/6/120

00/00/

/2/60/4/60

/6/120/6/120

00/00/

'

22

2323

22

2323

1

3

4

6

2

5

1 2 3 4 5 6

[ k] = [ T]T[ k ][T] =

Ui

Vj Mj

- iy6EI

L2

Ui Vi Mi

- iy6EI

L2

Uj

AE

L- ixiy)(

12EI

L3

AE

Lixjx

+12EI

L3iyjy)-(

AE

Lixjy

-12EI

L3iyjx)-()(

AE

Lix

2 +12EI

L3iy

2


130/148

130

Vi

Mi

Uj

Vj

Mj

ix6EI

L2

2EI

L

jy6EI

L2

- jx6EI

L2

4EI

L

ix6EI

L2

4EI

L

jy6EI

L2

jx6EI

L2-

2EI

L

AE

Liy

2 +12EI

L3ix

2)(

ix6EI

L2

ix6EI

L2

AE

L

iyjx-

12EI

L3ixjy)-(

jy6EI

L2

AE

L

jx2 +

12EI

L3

jy2)(

jy6EI

L2

jx6EI

L2-

AE

L

- jxjy)(12EI

L3

- jx6EI

L2

AE

L- ixiy)(

12EI

L3

- iy6EI

L2

- iy6EI

L2

AE

L

ixjx+

12EI

L3iyjy)-(

AE

Liyjy

+12EI

L3ixjx )-(

AE

Liyjy

+12EI

L3ixjx)-(

12EI

L3jx

2)

AE

L- ( ixjy- iyjx)

12EI

L3

AE

Liyjx

-12EI

L3ixjy)-(

AE

L- jxjy)(

12EI

L3

AE

Ljy

2 +(

Example 12

For the beam shown:

(a) Use the stiffness method to determine all the reactions at supports.


(c) Draw the quantitative bending moment diagrams


131/148

131

40 kN

4 m4 m 22.02 o



TakeI= 200(10

6

) mm

4

, A = 6(10

3

) mm

2

, andE= 200 GPa for all members.Include axial deformation in the stiffness matrix.

40 kN

4 m4 m 22.02o


132/148

132

40 kN40 kNm

20 kN

40 kNm

20 kN

[ qF ]

[FEF]

1Global

i j

1

i j

Local

20sin22.02=7.5

20cos22.02=18.54

1

2

3

4

5

6

4

5

6

x*

x1*

2*3*

ix = cos 0o = 1,

iy= cos 90o = 0 jx= cos 22.02o = 0.9271,jy = cos 67.98

o = 0.3749

22.02 o


133/148

1

i j1

2

3

4

5

6

Local

Local stiffness matrix

i j ji ji


134/148

134

[k]1 = 103

1

23

4

5

6

1 4 5 62 3

-150.0

0.000

0.000

150.0

0.000

0.000

0.000

-0.9375

-3.750

0.000

0.9375

-3.750

0.000

3.750

10.00

0.000

-3.750

20.00

150.0

0.000

0.000

-150.0

0.000

0.000

0.000

0.9375

3.750

0.000

-0.9375

3.750

0.000

3.750

20.00

0.000

-3.750

10.00

[ ]

=

LEILEILEILEI

LEILEILEILEI

LAELAE

LEILEILEILEILEILEILEILEI

LAELAE

k

/4/60/2/60

/6/120/6/120

00/00/

/2/60/4/60/6/120/6/120

00/00/

22

2323

22

2323

66

Mi

Vj

Mj

Vi

Nj

Ni

Stiffness matrix [k]:

1

1 4 5 62 3

-150.0 0.000 0.000150.0 0.000 0.000

x*

x1

4

5

6

1*

2*3* 1

i j1

2

3

4

5

6

Global Local

4

5

6

2*


135/148

135

[k]1 = 103

1

2

3

4

5

6

0.000

0.000

150.0

0.000

0.000

-0.9375

-3.750

0.000

0.9375

-3.750

3.750

10.00

0.000

-3.750

20.00

0.000

0.000

-150.0

0.000

0.000

0.9375

3.750

0.000

-0.9375

3.750

3.750

20.00

0.000

-3.750

10.00

Stiffness matrix [k*]: [ k* ] = [ T]T[ k ][T]

[k*]1 = 103

4

56

1*

2*

3*

4 1* 2* 3*5 6

-139.0

0.351

1.406

129.0

1.406

51.82

-56.25

-0.869

-3.750

51.82

21.90

-3.476

0.000

3.750

10.00

1.406

-3.476

20.00

150.0

0.000

0.000

-139.0

-56.25

0.000

0.000

0.9375

3.750

0.351

-0.869

3.750

0.000

3.750

20.00

1.406

-3.750

10.00

x*

x1

4

5

6

1*

2*3*

4 m4 m 22.02 o

40 kN

x*

x4

5

6

2*

40 kN40 kNm

40 kNm [ F ]


136/148

136

Global Equilibrium:

Q1 = 0.0

Q3 = 0.0

D1*

D3*=

36.37x10-6 m

0.002 rad

[Q] = [K][D] + [QF]

= 1031*

3*

1*

129

1.406

3*

1.406

20.0

D1*

D3*

-7.5

-40+

20 kN

40 kNm

20 kN

[ qF ]

20sin22.02=7.5

20cos22.02=18.54

4 m4 m 22.02 o

40 kN

x*

x1

4

5

6

1*

2*3*

40 kN 40 kNm40 kNm

4

5

6

2*


137/148

137

Member Force : [q] = [k*][D] + [qF]

0

40.0

20

-7.54

18.54

-40.0

+

q4

q6

q5

q1*

q2*q3*

= 103

4

5

6

1*

2*

3*

4 1* 2* 3*5 6

-139.0

0.351

1.406

129.0

1.406

51.82

-56.25

-0.869

-3.750

51.82

21.90

-3.476

0.000

3.750

10.00

1.406

-3.476

20.00

150.0

0.000

0.000

-139.0

-56.25

0.000

0.000

0.9375

3.750

0.351

-0.869

3.750

0.000

3.750

20.00

1.406

-3.750

10.00

0

0

0

36.4x10-6

02x10-3

-5.06

60

27.5

0

13.48

0

=

40 kN

13.48 kN

60 kNm

27.5 kN

5.06 kN

7.518.5420 kN

4 m4 m 22.02 o

40 kN

5.05 kN27.51 kN

60.05 kNm

13.47 kN

40 kN

50 04


138/148

138

+

Bending moment diagram(kNm)

Deflected shape

50.04

-

-60.05

0.002 rad

Example 13

For the beam shown:

(a) Use the stiffness method to determine all the reactions at supports.



d lit ti d fl t d h


139/148

139

40 kN

22.02o

8 m 4 m 4 m

2EI, 2AE EI,AE

6 kN/m


TakeI= 200(106) mm4, A = 6(103) mm2, andE= 200 GPa for all members.

Include axial deformation in the stiffness matrix.

40 kN

22.02o

8 m 4 m 4 m

2EI, 2AE EI,AE

6 kN/m

mkN

m

mkNm

L

AE

=

=

3

262

10150

)8(

)/10200)(006.0(


140/148

140

1 21

2

3

4

5

6

7*

8*9*

Global

21

2

3

4

56

11

2

3

4

5

6

Local

mkN10150

mkN

mmmkN

LEI

=

=

3

426

1020

)8()0002.0)(/10200(44

mkNL

EI

=3

1010

2

kN

m

mmkN

L

EI

3

2

426

2

1075.3

)8(

)0002.0)(/10200(66

=

=

mkN

m

mmkN

L

EI

/109375.0

)8(

)0002.0)(/10200(1212

3

3

426

2

=

=

1 21

2

3

4

5

6

7*

8*9*

GlobalLocal : Member 1

6 3 3 6


141/148

141

[k]1

= [k]1

= 2x103

4

5

6

12

3

4 1 2 35 6

-150.0

0.000

0.000

150.0

0.000

0.000

0.000

-0.9375

-3.750

0.000

0.9375

-3.750

0.000

3.750

10.00

0.000

-3.750

20.00

150.0

0.000

0.000

-150.0

0.000

0.000

0.000

0.9375

3.750

0.000

-0.9375

3.750

0.000

3.750

20.00

0.000

-3.750

10.00

14

5

1

2

[ q

]11

2

4

5

[ q]

[q] = [q]

Thus, [k] = [k]

1 21

2

3

4

5

6

7*

8*9*

23

x7 *

8*9*[ q* ] 2

3

4

56

[ q ]


142/148

142

Member 2: Use transformation matrix, [q*] = [T]T[q]

q1

q3

q2

q4

q5

q6

q1

q3

q2

q7*

q8*q9*

0.9271

0.3749

-0.3749

0.9271

0

0

0

0

0

0

0

0

0

0 0

0

0

1

1

0

0

1

=

1

2

3

7*

8*

9*

1 4 5 62 3

0 0

0

0

1

0

0

0

0

0

0

0

0

0

ix = cos 0o = 1,iy = cos 90

o = 0 jx = cos 22.02o = 0.9271,

jy = cos 67.98o = 0.3749

21

x*

7*21 4

Stiffness matrix [k]:

1

1 4 5 62 3

-150.0 0.000 0.000150.0 0.000 0.000

21

23

x*

x7*

8*9*[ q* ]

21

2

3

4

56

[ q ]


143/148

143

[k]2 = 103

2

3

4

5

6

0.000

0.000150.0

0.000

0.000

-0.9375

-3.7500.000

0.9375

-3.750

3.750

10.000.000

-3.750

20.00

0.000

0.000

-150.0

0.000

0.000

0.9375

3.7500.000

-0.9375

3.750

3.750

20.000.000

-3.750

10.00

Stiffness matrix [k*]: [k*] = [T]T[k][T]

[k*]2 = 103

1

2

3

7*

8*

9*

1 7* 8* 9*2 3

-139.0

0.351

1.406

129.0

1.406

51.82

-56.25

-0.869

-3.477

51.82

21.90

-3.476

0.000

3.750

10.00

1.406

-3.476

20.00

150.0

0.000

0.000

-139.0

-56.25

0.000

0.000

0.9375

3.750

0.351

-0.869

3.750

0.000

3.750

20.00

1.406

-3.477

10.00

1 21

2

3

4

5

6

7*

8*9*

1

2

3

4

5

6

8*

4

5

4 1 2 35 6

-150.0

0 000

0.000

0 9375

0.000

3 750

150.0

0 000

0.000

0 9375

0.000

3 750


144/148

144

[k]1= 2x103

5

61

2

3

0.000

0.000150.0

0.000

0.000

-0.9375

-3.7500.000

0.9375

-3.750

3.750

10.000.000

-3.750

20.00

0.000

0.000

-150.0

0.000

0.000

0.9375

3.7500.000

-0.9375

3.750

3.750

20.000.000

-3.750

10.00

[k*]2 = 103

1

2

37*

8*

9*

1 7* 8* 9*2 3

-139.0

0.351

1.406

129.0

1.406

51.82

-56.25

-0.869

-3.477

51.82

21.90

-3.476

0.000

3.750

10.00

1.406

-3.476

20.00

150.0

0.000

0.000

-139.0

-56.25

0.000

0.000

0.9375

3.750

0.351

-0.869

3.750

0.000

3.750

20.00

1.406

-3.477

10.00

32 kNm32 kNm

6 kN/m

40 kN6 kN/m

1 21

2

3

4

5

6

7*

8*

9*

1

2

3

4

5

6

8*

40 kN40 kNm

40 kNm40 kNm

40 kNm32 kNm


145/148

145

Global:

=

-509.84x10-6 rad

18.15x10-6 m

0.00225 rad

58.73x10-6 m

D3

D1

D9*

D7*

0

0

0

0 +-32 + 40 = 8

0

-40

-7.5

D3

D1

D9*

D7*

24 kN24 kN

1

= 1030

0

-139

450

10

1.40660

0

1.406

1.406

-139

129

0

101.406

20

1 3 7* 9*

1

37*

9*

7.5

18.5420 kN

7.5

Member 1: [ q] = [k][d] + [qF]

4 1 2 35 6

32 kNm

24 kN

32 kNm

24 kN

6 kN/m

114

5

6

1

2

3


146/148

146

0

32

24

0

24-32

+

0

0

0

d1=18.15x10-6

0d3=-509.84

x10-6

q4

q6

q5

q1

q2q3

= 2x103

45

6

1

23

4 1 2 35 6

-150.0

0.000

0.000

150.0

0.0000.000

0.000

-0.9375

-3.750

0.000

0.9375-3.750

0.000

3.750

10.00

0.000

-3.75020.00

150.0

0.000

0.000

-150.0

0.0000.000

0.000

0.9375

3.750

0.000

-0.93753.750

0.000

3.750

20.00

0.000

-3.75010.00

-5.45 kN

21.80 kNm

20.18 kN

5.45 kN

27.82 kN

-52.39 kNm

=

q4

q6

q5

q1

q2q3

6 kN/m

5.45 kN

20.18 kN

21.80 kNm5.45 kN

27.82 kN

52.39 kNm

21

3

x*

x7*

9*

1 7* 8* 9*2 3

20sin22.02=7.5

20cos22.02=18.54

40 kN40 kNm

20 kN

40 kNm

20 kN

[ qF ]

2

2 8*

Member 2: [ q] = [k][d] + [qF]


147/148

147

q1

q3

q2

q7*

q8*q9*

=103

12

3

7*

8*

9*

-139.0

0.351

1.406

129.0

1.40651.82

-56.25

-0.869

-3.750

51.82

21.90-3.476

0.000

3.750

10.00

1.406

-3.47620.00

150.0

0.000

0.000

-139.0

-56.250.000

0.000

0.9375

3.750

0.351

-0.8693.750

0.000

3.750

20.00

1.406

-3.75010.00

0

40

20

-7.5

18.54

-40

+

18.15x10-6

-509.84x10-6

0

58.73x10-6

0

0.00225

q1

q3

q2

q7*

q8*q9*

-5.45 kN

52.39 kNm

26.55 kN

0 kN

14.51 kN

0 kNm

= 5.45 kN

26.55 kN

52.39 kNm

14.51 kN

40 kN

5.45 kN

26.55 kN

52.39 kNm

14.51 kN

40 kN6 kN/m

5.45 kN

27.82 kN

52.39 kNm

5.45 kN

20.18 kN

21.80 kNm

40 kN6 kN/m21.80 kNm

14.51cos 22.02o

=13.45 kN


148/148

148

22.02o

8 m 4 m 4 m

5.45 kN

20.18 kN 14.51 kN54.37 kN

3.36 m

M

(kNm) x (m)

53.81

-21.8

-

V(kN)

x (m)

26.55

+

20.18

+- -

-27.82

-52.39

-

12.14 +

-13.45

Documents

06 Matrix Beam