stiffness 5

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Lecture No. : 5 الخامسة المحاضرة

Remember :F1

1 2

k1k2

F3

3

k3

F2

d1 d2 d3

k11F1

F2 = k21

F = K D

F3 k31

k12

k22

k32

k13

k23

k33

d1

d2

d32

k11F1

F2 = k21

F3 k31

k12

k22

k32

k13

k23

k33

d1

d2

d3

First column in Stiffness matrix

d1 =1 d2 =0 d3 =0

1 2

k1k2

3

k3

d1=1

k11

k21

k31

=

F1

F2

F3

Remember :

3

k11F1

F2 = k21

F3 k31

k12

k22

k32

k13

k23

k33

d1

d2

d3

Second column in Stiffness matrix

d1 =0 d2 =1 d3 =0

k12

k22

k32

=

F1

F2

F3

1 2

k1k2

3

k3

d2=1

Remember :

4

k11F1

F2 = k21

F3 k31

k12

k22

k32

k13

k23

k33

d1

d2

d3

Third column in Stiffness matrix

d1 =0 d2 =0 d3 =1

k13

k23

k33

=

F1

F2

F3

1 2

k1k2

3

k3

d3=1

Remember :

5

Plan Truss

A

B

C

4

3

F1d1

F2 d2

F = K Dk11F1

F2

=k21

k12

k22

d1

d2

Remember :

6

First column in Stiffness matrixd1 =1 d2 =0

k11F1

F2

=k21

k12

k22

d1

d2

A

B

C

4

3

d1

1

Remember :

7

second column in Stiffness matrixd1 =0 d2 =1

k11F1

F2

=k21

k12

k22

d1

d2

A

B

C

d2

1

Remember :

8

a

b

F1

d1=1

F2

F3

F4

= EA/L cos2

= EA/L cos sin

= - EA/L cos2

= - EA/L cos sin

9

a

b

F1

F2

F3

F4

= EA/L sin2

= EA/L sin cos

= - EA/L sin cos

= - EA/L sin2

d2=1

10

( )cos + ( )sin

bxax

- byay

-AEL

N =

a

b

bx

by

ax

ay

[ ]

For internal forces in truss elementsRemember :

11

4

A

B

E

5 2 4C D

B

Beams

C D

Remember :

12

k11

K = k21

k31

k12

k22

k32

k13

k23

k33

Stiffness matrix

Remember :

13

A

B

E

C D

B

First column in Stiffness matrix

B =1

4 EILAB

4 EILBC

+ 2 EILBC

Remember :

14

A

B

E

C D

C

Second column in Stiffness matrix

C =1

4 EILBC

4 EILCD

+ 2 EILCD

2 EILBC

Remember :

15

A

B

E

C D

D

Third column in Stiffness matrix

D =1

4 EILCD

4 EILDE

+2 EILCD

Remember :

16

A

B

E

C D

60 kNm 50 kNm 30 kNmForce vector

F =

- 60- 50

30

A

B

E

C D

B C D

Remember :

17

Force vectorTransformation from member forces to Joint forces

L

P

8LP

8LP

P a b2

L2

L

P

a b P b a2

L2

Fixed End Reaction (FER) 18

Force vectorTransformation from member forces to Joint forces

L

P

8LP

8LP

P a b2

L2

L

P

a b P b a2

L2

Fixed End Action (FEA) 19

Internal forces in beam elements

MBA=2 EI

L( + 2 ) M(FER) BA +

MAB= ( 2 + ) M(FER) AB +2 EI

L

20

21

I I

2 I

22

d1

d2

d3

23

d1

d2

d3

d11

d4

d5

d6d7

d8

d9

d10 d12

24

25

26

AEL

AEL

AEL

AEL

27

AEL

AEL

AEL

AEL

28

29

6 EI

L2

6 EIL2

12 EIL3

12 EIL3

30

6 EIL2

6 EIL2

12 EIL3

12 EIL3

31

3 EIL2

3 EIL3

3 EIL3

32

3 EIL2

3 EIL3

3 EIL3

33

34

4 EIL

2 EIL

6 EIL2

6 EIL2

35

4 EIL

2 EIL

6 EIL2

6 EIL2

36

3 EIL 3 EI

L2

3 EIL2

37

3 EIL

3 EIL2

3 EIL2

38

Example 1:Construct the stiffness matrix for the shown frame where E = 106 kN/m2

8

A B

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

39

Modeling

A

B

C

d1

d2

d3

40

k11

K = k21

k31

k12

k22

k32

k13

k23

k33

Stiffness matrix

41

First column in Stiffness matrix

d1 =1A

B

C

d1

42

First column in Stiffness matrix

A B

C

1 AEL 1 6 EI

L2

12 EIL3

B

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

43

First column in Stiffness matrix106x.6

8 6X106X.00552

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

12X106X.00553

75,0001200

480

44

First column in Stiffness matrix

k11

k21

k31

=

75,480

12000

A B

C

75,0001200

480d1

d2

d3

45

Second column in Stiffness matrix

d2 =1A

B

C

d2

46

A B

C

AEL

6 EIL2

12 EIL3

Second column in Stiffness matrix

11

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

47

A B

C

106x.45

6X106X.0282

12X106X.0283

80,000

1875469

Second column in Stiffness matrix

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

48

A B

C

80,000

1875469

k12

k22

k32

=

080,469

-1875

Second column in Stiffness matrix

d1

d2

d3

49

Third column in Stiffness matrix

d3 =1A

B

C

d3

50

Third column in Stiffness matrix

A

B

C

1

1

4 EIL

6 EIL2

4 EIL

6 EIL2

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

51

Third column in Stiffness matrix

A

B

C

1

1

6X106X.0282

4X106X.028

6X106X.00552

4X106X.0055

10,0004,000

1200

1875A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

52

Third column in Stiffness matrix

A

B

C

10,0004,000

1200

1875

d1

d2

d3

k13

k23

k33

=

1200-1875

14,00053

k11

k21

k31

=

75,480

1200

0k12

k22

k32

=

080,469

-1875

k13

k23

k33

=

1200-1875

14,000

K =

75,480

1200

00

80,469

-1875

1200-1875

14,000

54

Example 2:Construct the stiffness matrix for the shown frame where E = 106 kN/m2, and the section of all members is rectangular with dimension 30 cm x 80 cm.

8a

b

8

d

c

55

Modeling

d1

d2

d3

a

b

d

c

d4

d5

d6

56

k11

K =

k21

k31

k12

k22

k32

k13

k23

k33

Stiffness matrix

k41

k51

k61

k42

k52

k62

k43

k53

k63

k14

k24

k34

k15

k25

k35

k16

k26

k36

k44

k54

k64

k45

k55

k65

k46

k56

k66

57

Example 2:

30 cm

80 cmEA= 240,000 kNEI = 12,800 kNm2

Construct the stiffness matrix for the shown frame where E = 106 kN/m2, and the section of all members is rectangular with dimension 30cm x 80 cm.

A = 0.24 m2

I = 0.0128 m4

58

8a

b

8

d

c

EA=240,000 kNEI = 12,800 kNm2

EA/L = 30,0002 EI/L = 3,2004 EI/L = 6,4006 EI/L2 = 1,20012EI/L3= 300

59

First column in Stiffness matrix

d1 =1d1

a

b

d

c

60

First column in Stiffness matrix

b c

a

1 AEL1 6 EI

L2

12 EIL3

b

AEL

d

c

EA/L = 30,0002 EI/L = 3,2004 EI/L = 6,4006 EI/L2 = 1,20012EI/L3= 300

30,000 30,0001,200

300

61

First column in Stiffness matrix

b c

d

c

30,000 30,0001,200

300

a

b

d

c

d1

d2

d3 d4

d5d6

k11

=

k21

k31

k41

k51

k61

30,3000

1,200-30,000

00

62

Second column in Stiffness matrix

d2 =1

d2

a

b

d

c

63

b c

a

AEL

6 EIL2

12 EIL3

Second column in Stiffness matrix

1

1

12 EIL3

6 EIL2

b

d

c

EA/L = 30,0002 EI/L = 3,2004 EI/L = 6,4006 EI/L2 = 1,20012EI/L3= 300

30,0001,200

300

1,200

300

64

Second column in Stiffness matrix

30,0001,200

300

1,200

300

a

b

d

c

d1

d2

d3 d4

d5d6

k12

=

k22

k32

k42

k52

k62

30,3000

1,200

-300

0

1,20065

Third column in Stiffness matrix

d3 =1d3

a

b

d

c

66

Third column in Stiffness matrix

a

1 6 EIL2

4 EIL

6 EIL2

6 EIL2

4 EIL

2 EIL

1

d

c

EA/L = 30,0002 EI/L = 3,2004 EI/L = 6,4006 EI/L2 = 1,20012EI/L3= 300

6,4006,400

3,200

1,2001,200

1,200b c

67

Third column in Stiffness matrix

ad

c

6,400 6,400

3,200

1,200

1,2001,200

b

a

b

d

c

d1

d2

d3 d4

d5d6

k13

=

k23

k33

k43

k53

k63

1,2001,200

12,800

-1,200

0

3,200 68

Fourth column in Stiffness matrix

d4 =1

d4

a

b

d

cc

69

b c

d

1 AEL 1 6 EI

L2

12 EIL3

c

Fourth column in Stiffness matrix

a

b

AEL

EA/L = 30,0002 EI/L = 3,2004 EI/L = 6,4006 EI/L2 = 1,20012EI/L3= 300

30,000 30,000 1,200

300

70

Fourth column in Stiffness matrix

a

b

30,000 30,0001,200

300

a

b

d

c

d1

d2

d3 d4

d5d6

k14

=

k24

k34

k44

k54

k64

0-30,000

0

0

30,300

1,200

71

Fifth column in Stiffness matrix

d5 =1

d5

a

b

d

c

72

b c

d

AEL

6 EIL2

12 EIL3

11

Fifth column in Stiffness matrix

a

b

c

6 EIL2

12 EIL3

EA/L = 30,0002 EI/L = 3,2004 EI/L = 6,4006 EI/L2 = 1,20012EI/L3= 300

30,000

1,200

1,200300

300

73

Fifth column in Stiffness matrix

a

b

30,000

1,200

1,200300

300

a

b

d

c

d1

d2

d3 d4

d5d6

k15

=

k25

k35

k45

k55

k65

-3000

-1,200

30,300

0

-1,20074

Sixth column in Stiffness matrix

d6 =1

d6

a

b

d

c

75

b c

d

1

1

4 EIL

6 EIL2

4 EIL

6 EIL2

Sixth column in Stiffness matrix

a

b

c

2 EIL

6 EIL2

EA/L = 30,0002 EI/L = 3,2004 EI/L = 6,4006 EI/L2 = 1,20012EI/L3= 300

3,200 6,4006,400

1,2001,200

1,200

76

b c

d

Sixth column in Stiffness matrix

a

bc

3,200 6,400

6,400

1,200 1,200 1,200

a

b

d

c

d1

d2

d3 d4

d5d6

k16

=

k26

k36

k46

k56

k66

1,2000

3,200

-1,200

1,200

12,80077

k11

=

k21

k31

k41

k51

k61

30,3000

1,200-30,000

00

k12

=

k22

k32

k42

k52

k62

30,3000

1,200

-300

0

1,200

k13

=

k23

k33

k43

k53

k63

1,2001,200

12,800

-1,200

0

3,200

k14

=

k24

k34

k44

k54

k64

0-30,000

0

0

30,300

1,200

k15

=

k25

k35

k45

k55

k65

-3000

-1,200

30,300

0

-1,200

k16

=

k26

k36

k46

k56

k66

1,2000

3,200

-1,200

1,200

12,80078

=K

30,3000

1,200-30,000

00

30,3000

1,200

-300

0

1,200

1,2001,200

12,800

-1,200

0

3,200

0-30,000

0

0

30,300

1,200

-3000

-1,200

30,300

0

-1,200

1,2000

3,200

-1,200

1,200

12,800

79

Example 3:Construct the stiffness matrix for the shown frame where E = 106 kN/m2

8

A B

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

80

Modeling

A

B

C

d1

d2

d3

First Model No DOF at C

81

k11

K = k21

k31

k12

k22

k32

k13

k23

k33

Stiffness matrix

82

First column in Stiffness matrix

d1 =1A

B

C

d1

83

First column in Stiffness matrix

A B

C

1 AEL 1 3 EI

L2

3 EIL3

B

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

84

First column in Stiffness matrix106x.6

8 3X106X.00552

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

3X106X.00553

75,000600

120

85

First column in Stiffness matrix

k11

k21

k31

=

75,120

6000

A B

C

75,000600

120d1

d2

d3

86

Second column in Stiffness matrix

d2 =1A

B

C

d2

87

A B

C

AEL

6 EIL2

12 EIL3

Second column in Stiffness matrix

11

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

88

A B

C

106x.45

6X106X.0282

12X106X.0283

80,000

1875469

Second column in Stiffness matrix

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

89

A B

C

80,000

1875469

k12

k22

k32

=

080,469

-1875

Second column in Stiffness matrix

d1

d2

d3

90

Third column in Stiffness matrix

d3 =1A

B

C

d3

91

Third column in Stiffness matrix

A

B

C

1

1

4 EIL

6 EIL2

3 EIL

3 EIL2

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

92

Third column in Stiffness matrix

A

B

C

1

1

6X106X.0282

4X106X.028

3X106X.00552

3X106X.0055

10,0003,000

600

1875A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

93

Third column in Stiffness matrix

A

B

C

10,0003,000

600

1875

d1

d2

d3

k13

k23

k33

=

600-1875

13,00094

k11

k21

k31

=

75,120

600

0k12

k22

k32

=

080,469

-1875

k13

k23

k33

=

600-1875

13,000

K =

75,120

600

00

80,469

-1875

600-1875

13,000

95

Modeling

A

B

C

d1

d2

d3

Second Model Take DOF at C

d4

96

k11

K =k21

k31

k12

k22

k32

k13

k23

k33

Stiffness matrix

k41 k42 k43

k14

k24

k34

k44

97

First column in Stiffness matrix

d1 =1A

B

C

d1

98

First column in Stiffness matrix

A B

C

1 AEL 1 6 EI

L2

12 EIL3

B

6 EIL2

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

99

First column in Stiffness matrix106x.6

8 6X106X.00552

12X106X.00553

75,0001200

480

6X106X.00552

1200

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

100

First column in Stiffness matrix

k11

k21

k31

=

75,480

12000

A B

C

75,0001200

480d1

d2

d3

d4

1200k41 1200

101

Second column in Stiffness matrix

d2 =1A

B

C

d2

102

A B

C

AEL

6 EIL2

12 EIL3

Second column in Stiffness matrix

11

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

103

A B

C

106x.45

6X106X.0282

12X106X.0283

80,000

1875469

Second column in Stiffness matrix

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

104

A B

C

80,000

1875469

k12

k22

k32

=

080,469

-1875

Second column in Stiffness matrix

d1

d2

d3

d4 k42 0105

Third column in Stiffness matrix

d3 =1A

B

C

d3

106

Third column in Stiffness matrix

A

B

C

1

1

4 EIL

6 EIL2

4 EIL

6 EIL2

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

2 EIL 107

Third column in Stiffness matrix

A

B

C

1

1

6X106X.0282

4X106X.028

6X106X.00552

4X106X.0055

10,0004,000

1200

1875A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

2X106X.0055 2,000

108

Third column in Stiffness matrix

A B

C

10,0004,000

1200

1875

k13

k23

k33

=

1200-1875

14,000

d1

d2

d3

d4 2,000k43 2,000

109

Fourth column in Stiffness matrix

d4 =1A

B

Cd4

110

A B

C

1

2 EIL

6 EIL2

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

4 EIL

Fourth column in Stiffness matrix

111

A

B

C

1

6X106X.00552

2X106X.00552,000

1200

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

4X106X.0055 4,000

Fourth column in Stiffness matrix

112

A B

C

2,000

1200

k14

k24

k34

=

12000

2,000

d1

d2

d3

d4 4,000k44 4,000

Fourth column in Stiffness matrix

113

K =

k11

k21

k31

=

75,480

12000

k41 1200

k12

k22

k32

=

080,469

-1875k42 0

k13

k23

k33

=

1200-1875

14,000k43 2,000

k14

k24

k34

=

12000

2,000k44 4,000

75,480

12000

1200

080,469

-18750

1200-1875

14,000

2,000

12000

2,0004,000

114

Example 4:Construct the stiffness matrix for the shown frame where E = 106 kN/m2

8

A B

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

115

Modeling

A

B

C

d1

d2

d3

d5

d4

116

k11

K =

k21

k31

k12

k22

k32

k13

k23

k33

Stiffness matrix

k41 k42 k43

k14

k24

k34

k44

k51 k52 k53 k54

k15

k25

k35

k45

k55

117

First column in Stiffness matrix

d1 =1A

B

C

d1

118

First column in Stiffness matrix

A B

C

1 AEL 1 6 EI

L2

12 EIL3

B

6 EIL2

12 EIL3

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

119

First column in Stiffness matrix106x.6

8 6X106X.00552

12X106X.00553

75,0001200

480

1200

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

480

120

First column in Stiffness matrix

k11

k21

k31 =

75,480

12000

A B

C

75,0001200

480

1200

k41 - 480

d1

d2

d3

d5

d4

480

k51 1200 121

Second column in Stiffness matrix

d2 =1A

B

C

d2

122

A B

C

AEL

6 EIL2

12 EIL3

Second column in Stiffness matrix

11

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

123

A B

C

106x.45

6X106X.0282

12X106X.0283

80,000

1875469

Second column in Stiffness matrix

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

124

AB

C

80,000

1875469

k12

k22

k32

=

080,469

-1875

Second column in Stiffness matrix

k42 0

d1

d2

d3

d5

d4

k52 0125

Third column in Stiffness matrix

d3 =1A

B

C

d3

126

Third column in Stiffness matrix

A

B

C

1

1

4 EIL

6 EIL2

4 EIL

6 EIL2

2 EIL

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2 6 EIL2

127

Third column in Stiffness matrix

A

B

C

1

1

6X106X.0282

4X106X.028

6X106X.00552

4X106X.0055

10,0004,000

1200

1875

2X106X.0055 2,000

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2 6X106X.00552

1200

128

Third column in Stiffness matrix

A B

C

10,0004,000

1200

1875k13

k23

k33

=

1200-1875

14,000

2,000

k43 -1,200

d1

d2

d3

d5

d4

1200

k53 2,000

129

Fourth column in Stiffness matrix

d4 =1A

B

Cd4

130

A B

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

Fourth column in Stiffness matrix

1

6 EIL212 EI

L3

6 EIL2

12 EIL3

131

A

B

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

Fourth column in Stiffness matrix6X106X.005

52

12X106X.00553

1200480

12X106X.00553

1200

480

6X106X.00552 132

A B

k14

k24

k34 =

- 4800

-1,200k44 480

d1

d2

d3

d5

d4

Fourth column in Stiffness matrix

480

1200

480

1200

k54 -1,200133

Fifth column in Stiffness matrix

d5 =1A

B

Cd5

134

A B

C

1

2 EIL

6 EIL2

4 EIL

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

Fifth column in Stiffness matrix

6 EIL2

135

A

B

C

1

6X106X.00552

2X106X.00552,000

1200

4X106X.0055 4,000

A B

8

5

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

Fifth column in Stiffness matrix

6X106X.00552 1200

136

A B

C

2,000

1200

k15

k25

k35 =

12000

2,000

4,000

k45 -1,200d1

d2

d3

d5

d4

Fifth column in Stiffness matrix

1200

k55 4,000

137

k11

k21

k31 =

75,480

12000

k41 - 480k51 1200

k12

k22

k32 =

080,469

-1875k42 0k52 0

k13

k23

k33 =

1200-1875

14,000k43 -1,200k53 2,000

k15

k25

k35 =

12000

2,000k45 -1,200

k55 4,000

k14

k24

k34 =k44

k54

- 4800

-1,200480

-1,200138

K =

75,480

12000

- 4801200

080,469

-187500

1200-1875

14,000-1,2002,000

12000

2,000

-1,200

4,000

- 4800

-1,200480

-1,200

139

Example 5:Construct the stiffness matrix for the shown frame where E = 106 kN/m2

8

A

B

6

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

140

Modeling

B

C

d1

d2

d3

A

141

k11

K = k21

k31

k12

k22

k32

k13

k23

k33

Stiffness matrix

142

First column in Stiffness matrix

d1 =1B

C

d1

A

143

First column in Stiffness matrix

C

1 6 EIL2

12 EIL3

B

A

B

8

6

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

144

First column in Stiffness matrix

6X106X.00562

12X106X.00563

833.33

277.78

A

B

8

6

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

145

First column in Stiffness matrixsin 6EI sin

L212EI sin

L3 B

A

1

sin

cos

EA cos L

146

First column in Stiffness matrix 6EI sinL2

12EI sinL3

B

A

EA cos L

147

First column in Stiffness matrix 6EI sinL2

12EI sinL3

BA

EA cos L

EA cos2

L

EA cos sin

L

12EI sin2L3

12EI sincos

L3

EA cos2 L

12EI sin2L3+

EA cos sin

L-12EI sincos

L3

148

First column in Stiffness matrix

6EI sinL2

B

A

EA cos2 L

12EI sin2L3+

EA cos sin

L-12EI sincos

L3

A

B

8

6

C

A=0.6 m2

I = 0.02 m

4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

106x.610

= 60,000AEL

=6 EI

L2

12 EIL3

6X106X.02102

12X106X.02103

1200

240

=

= =

=

cos = 0.8sin = 0.6

149

First column in Stiffness matrix

6EI sinL2

B

A

EA cos2 L

12EI sin2L3+

EA cos sin

L-12EI sincos

L3

cos = 0.8sin = 0.6

= 60,000AEL

6 EIL2

12 EIL3

= 1200 = 240

60,000x0.64+240x0.36

60,000x0.48- 240x0.48

38,486.4

28,684.8

1200x0.6 720

150

First column in Stiffness matrix

B

A

38,486.428,684.8

720

38,764.18

28,684.8

1553.33

k11

k21

k31

=

38,764.18

1553.3328,684.8

833.33

277.78

151

Second column in Stiffness matrix

d2 =1

B

C

d2

A

152

C

AEL

Second column in Stiffness matrix

1

106x.46

A

B

8

6

C

A=0.6 m2

I = 0.02 m

4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

66,666.67

153

6EI cosL2

12EI cosL3

B

A

1

cos

sin EA sin

Lcos sin

Second column in Stiffness matrix

154

6EI cosL2

12EI cosL3

B

A

EA sin L

Second column in Stiffness matrix

155

6EI cosL2

12EI cosL3

BA

EA sin L

EA sin2 L

EA sin cos

L

12EI cos2L3

12EI cossin

L3

EA sin2 L

12EI cos2L3+

EA sincos

L-12EI cossin

L3

Second column in Stiffness matrix

156

6EI cosL2

B

A A

B

8

6

C

A=0.6 m2

I = 0.02 m

4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

106x.610

= 60,000AEL

=6 EI

L2

12 EIL3

6X106X.02102

12X106X.02103

1200

240

=

= =

=

EA sin2 L

12EI cos2L3+

EA sincos

L-12EI cossin

L3

Second column in Stiffness matrix

cos = 0.8sin = 0.6

157

6EI cosL2

B

A

cos = 0.8sin = 0.6

= 60,000AEL

6 EIL2

12 EIL3

= 1200 = 240

60,000x0.48-240x0.48

60,000x0.36+240x0.64

28,684.8

21,753.6

1200x0.8 960

EA sin2 L

12EI cos2L3+

EA sincos

L-12EI cossin

L3

Second column in Stiffness matrix

158

B

A

28,684.8

88,420.27

960

k12

k22

k32

=

28,684.8

- 96088,420.27

28,684.821,753.6

960

Second column in Stiffness matrix

C

66,666.67

159

Third column in Stiffness matrix

d3 =1

B

C

d3

A

160

Third column in Stiffness matrix

C

1A

B 1

4 EIL

6 EIL2

4 EIL

6 EIL2

161

Third column in Stiffness matrix

C

1

4 EIL

6 EIL2

A

B

8

6

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

6X106X.00562

4X106X.00563,333.33

833.33

162

Third column in Stiffness matrix

A

B

4 EIL

6 EIL2

163

Third column in Stiffness matrix

A

4 EIL

6 EIL2

6 EI sin L2

6 EI cos L2

164

Third column in Stiffness matrix

A

4 EIL

6 EI sin L2

6 EI cos L2

A

B

8

6

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

4X106X.0210

6X106X.02102

cos = 0.8sin = 0.6

6 EIL2

= 1,200=

720

960

8,000

165

Third column in Stiffness matrixA

B8,000

C

3,333.33

833.33960

720

k13

k23

k33

=

1553.33-960

11,333.33

11,333.33

960

1553.33

166

K =

k13

k23

k33

=

1553.33

-960

11,333.33

k12

k22

k32

=

28,684.8

- 960

88,420.27

k11

k21

k31

=

38,764.18

1553.33

28,684.8

1553.33

-960

11,333.33

28,684.8

- 960

88,420.27

38,764.18

1553.33

28,684.8

167

168

Bd4

d5

d6

A

d1

d2

d3

169

sin 6EI sin

L212EI sin

L3 B

A

1

sin

cos

EA cos L

Horizontal Deformation

170

6EI sinL2

12EI sinL3

B

A

EA cos L

171

6EI sinL2

12EI sinL3

BA

EA cos L

EA cos2

L

EA cos sin

L

12EI sin2L3

12EI sincos

L3

EA cos2 L

12EI sin2L3+

EA cos sin

L-12EI sincos

L3

172

6EI sinL2

EA cos2 L

12EI sin2L3+

EA cos sin

L-12EI sincos

L3

6EI sinL2

6EI sinL2

EA cos2 L

12EI sin2L3+

EA cos sin

L-12EI sincos

L3

173

6EI cosL2

12EI cosL3

B

A

1

cos

sin EA sin

Lcos sin

Vertical Deformation

174

6EI cosL2

12EI cosL3

B

A

EA sin L

175

6EI cosL2

12EI cosL3

BA

EA sin L

EA sin2 L

EA sin cos

L

12EI cos2L3

12EI cossin

L3

EA sin2 L

12EI cos2L3+

EA sincos

L-12EI cossin

L3 176

6EI cosL2

EA sin2 L

12EI cos2L3+

EA sincos

L-12EI cossin

L3

6EI cosL2

EA sincos

L-12EI cossin

L3

EA sin2 L

12EI cos2L3+

6EI cosL2

-177

1

4 EIL

6 EIL2

2 EIL

Rotational Deformation

178

A

B

4 EIL

6 EIL2

2 EIL

179

4 EIL

6 EIL2

6 EI sin L2

6 EI cos L2

2 EIL

180

4 EIL

6 EI sin L2

6 EI cos L2

2 EIL

-

6 EI sin L2

6 EI cos L2

4 EIL

181

Example 6:Construct the stiffness matrix for the shown frame where E = 106 kN/m2, A = 0.6 m2, I = .02 m4

A B

O

5

C D E

150 135 60 45

182

Modeling

A B

O

5

C D E

150 135 60 45

d1

d2

d3

183

k11

K = k21

k31

k12

k22

k32

k13

k23

k33

Stiffness matrix

184

6EI sinL2

EA cos2 L

12EI sin2L3+

EA cos sin

L-12EI sincos

L3

Horizontal Deformation

185

EA sincos

L-12EI cossin

L3

EA sin2 L

12EI cos2L3+

6EI cosL2

-

Vertical Deformation

186

-

6 EI sin L2

6 EI cos L2

4 EIL

Rotational Deformation

187

EA cos2 L

12EI sin2

L3+

EA cossin

L-12EI sincos

L3

6EI sin

L2

EA sincos

L-12EI cossin

L3

EA sin2 L

12EI cos2

L3+

6EI cos

L2-

6EI sin

L2

6EI cos

L2-

4 EI

L

K =

188

member-

-

L--

EA/L-

4EI/L-

6 EI / L2

_12 EI / L3

_

OA 150 10.02 59890.5 7985.4 1195.6 238.7

OB 135 7.08 84772.3 11303.0 2395.4 676.9

OC 90 5.00 120000.0 16000.0 4800.0 1920.0

OD 60 5.77 103948.3 13859.8 3601.8 1248.0

OE 45 7.07 84879.6 11317.3 2401.5 679.5

A B

O

5

C D E

150 135 60 45

189

Member-

cos

sin

cos2

sin2

sin cos

OA 150 -0.867 0.499 0.751 0.249 -0.432

OB 135 -0.708 0.706 0.501 0.499 -0.500

OC 90 -0.001 1.000 0.000 1.000 -0.001

OD 60 0.500 0.866 0.250 0.750 0.433

OE 45 0.707 0.707 0.500 0.500 0.500

A B

O

5

C D E

150 135 60 45

190

EA cos2 L

12EI sin2

L3+

EA cossin

L-12EI sincos

L3

6EI sin

L2

EA sincos

L-12EI cossin

L3

EA sin2 L

12EI cos2

L3+

6EI cos

L2-

6EI sin

L2

6EI cos

L2-

4 EI

L

k11

k21

k31

k22

k32 k33

191

member k11 k21 k31 k22 k32 k33

OA 45031.92 -25798.5 297.8152 15097.23 1036.069 7985.394

OB 42804.35 -42047.6 1195.452 42644.84 1695.443 11302.97

OC 1920.046 -74.6554 4799.996 119999.9 3.034773 16000

OD 26885.59 44448.89 2702.628 78310.72 -1799.56 13859.78

OE 42752.93 42100.08 1201.518 42806.17 -1697.59 11317.28

159394.8 18628.19 10197.41 298858.9 -762.607 60465.43

159394.8

K = 18628.19

10197.41

18628.19

298858.9

762.607

10197.41

762.607

60465.43

192

( )cos + ( )sin

bxax

- byay

-AEL

N =

a

b

bx

by

ax

ay

[ ]

Internal forces

Normal Force

193

Internal forces

Bending momentDue to rotational deformation

MBA=2 EI

L( + 2 ) M(FER) BA +

MAB= ( 2 + ) M(FER) AB +2 EI

L

194

a

b

bx

by

ax

ay

( )cos

- ( )sin

bxax

-byay

- =

Internal forces

Bending momentDue to shear deformation

6EI L2

6EI

L2

195

Internal forces

Bending moment

MBA=2 EI

L( + 2 )

M(FER) BA +

MAB= ( 2 + ) M(FER) AB +2 EI

L6EI

L2-

6EI L2

-

MAB= ( 2 + ) M(FER) AB +2 EI

L3 L-

MBA=2 EI

L( + 2 )

M(FER) BA +3 L-

196

Example 7:Draw N.F, S.F & B.M.Ds for the shown frame where E = 106 kN/m2

8

A B

3

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m450 kN

12 kN/m

2

197

Modeling

A

B

C

d1

d2

d3

d5

d4

198

K =

75,480

12000

- 4801200

080,469

-187500

1200-1875

14,000-1,2002,000

12000

2,000

-1,200

4,000

- 4800

-1,200480

-1,200

From Example 4 :

199

Force vector

8

A B

C

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m450 kN

12 kN/m

3

2

200

Force vector

8

A

12 kN/m

B

64 kNm12x82

12 64 kNm

48 kN 48 kN

201

Force vector

B

C

50 kN

3

2

24 kNm

50x2x32

52

36 kNm

32.4 kN

50x3x22

52

17.6 kN

202

Force vector

A B

64 kNm 64 kNm

48 kN 48 kN B

C

24 kNm

36 kNm

32.4 kN

17.6 kN

Fixed End Reaction

(FER)

203

Force vector

A B

64 kNm 64 kNm

48 kN 48 kN B

C

24 kNm

36 kNm

32.4 kN

17.6 kN

Fixed End Action (FEA)

AB

C

d1

d2

d3

d5 d4

F =

F1 -17.6

=

F2 - 48F3 40F4 -32.4

F5 36204

F = K D

d1

d2

-17.6

=- 4840

-32.4

36

75,480

12000

- 4801200

080,469

-187500

1200-1875

14,000-1,2002,000

12000

2,000

-1,200

4,000

- 4800

-1,200480

-1,200

d3

d4

d5

205

D = K-1 F

d1

d2

-17.6

=- 4840

-32.4

36

75,480

12000

- 4801200

080,469

-187500

1200-1875

14,000-1,2002,000

12000

2,000

-1,200

4,000

- 4800

-1,200480

-1,200

d3

d4

d5

-1

206

D = K-1 F

d1

d2

- 0.0007

=- 0.0008- 0.0088- 0.2244

- 0.0538

d3

d4

d5

207

Internal forcesd1d2

- 0.0007

=

- 0.0008

- 0.0088

- 0.2244

- 0.0538

d3

d4

d5

AB

C

d1

d2

d3

d5 d4

Bending moment

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

EIAB = 2x104

EIBC = 5x103

208

Force vector

A B

64 kNm 64 kNm

48 kN 48 kN B

C

24 kNm

36 kNm

32.4 kN

17.6 kN

Fixed End Reaction

(FER)

209

= 64 + 2x2x104/8 (-.0088-3x-.0008/8) = 21.75

A B

64 kNm 64 kNm

48 kN 48 kN

d1d2

- 0.0007

=

- 0.0008

- 0.0088

- 0.2244

- 0.0538

d3

d4

d5

MAB= ( 2 + ) M(FER) AB +2 EI

L3 L-

MBA=2 EI

L( + 2 )

M(FER) BA +3 L-

= -64 + 2x2x104/8 (2x-.0088-3x-.0008/8) = -150

d1

d2d3

d5d4

210

= 24 + 2x5x103/5 (2x-.0088-.0538-3x-.2237/5) = 150

d1d2

- 0.0007

=

- 0.0008

- 0.0088

- 0.2244

- 0.0538

d3

d4

d5

MBC= ( 2 + )B CM(FER) AB +2 EI

L3 L-

MCB=2 EI

L( + 2 )B CM(FER) BA +

3 L-

= -36 + 2x5x103/5 (-.0088+2x-.0538-3x-.2237/5) = 0

B

C

24 kNm

36 kNm

17.6 kN

=d4- d1=-0.2237

d1

d2d3

d5d4

211

A

12 kN/m

B

21.75 kNm 150 kNm

48 kN 48 kN

16 kN 16 kN

32 kN 64 kN

B

C

50 kN

150 kNm

0 kN

50 kN3

28

212

Internal forcesd1d2

- 0.0007

=

- 0.0008

- 0.0088

- 0.2244

- 0.0538

d3

d4

d5

AB

C

d1

d2

d3

d5 d4

A=0.6 m2

I = 0.02 m4

A=0.4 m2

I = 0.005 m4

E = 106 kN/m2

EAAB = 6x105

EABC = 4x105

Normal Force AEL

N =

NAB = 6x105 /8 x -.0007 = - 50NBC = 4x105 /5 x -.0008 = - 64

213

A B 64 kN

B

C

50 kN

214

A

12 kN/m

B

21.75 kNm 150 kNm

32 kN 64 kN B

C

50 kN

150 kNm

50 kN

64 kN50 kN

215

A

12 kN/m

B

21.75 kNm 150 kNm

32 kN 64 kN B

C

50 kN

150 kNm

50 kN

64 kN50 kN

N.F.D

50

64

-

-

216

A

12 kN/m

B

21.75 kNm 150 kNm

32 kN 64 kN B

C

50 kN

150 kNm

50 kN

64 kN50 kN

S.F.D

64

50

50

32

-

+

+

217

A

12 kN/m

B

21.75 kNm 150 kNm

32 kN 64 kNB

C

50 kN

150 kNm

50 kN

64 kN

50 kN

B.M.D

-

15021.75

15085.875

96

10.125

--+

218

Summary

219

d1

d2

d3

220

AEL

AEL

AEL

AEL

221

6 EI

L2

6 EIL2

12 EIL3

12 EIL3

222

6 EIL2

6 EIL2

12 EIL3

12 EIL3

223

3 EIL2

3 EIL3

3 EIL3

224

3 EIL2

3 EIL3

3 EIL3

225

4 EIL

2 EIL

6 EIL2

6 EIL2

226

4 EIL

2 EIL

6 EIL2

6 EIL2

227

3 EIL 3 EI

L2

3 EIL2

228

3 EIL

3 EIL2

3 EIL2

229

6EI sinL2

EA cos2 L

12EI sin2L3+

EA cos sin

L-12EI sincos

L3

6EI sinL2

6EI sinL2

EA cos2 L

12EI sin2L3+

EA cos sin

L-12EI sincos

L3

Horizontal Deformation

230

6EI cosL2

EA sin2 L

12EI cos2L3+

EA sincos

L-12EI cossin

L3

6EI cosL2

EA sincos

L-12EI cossin

L3

EA sin2 L

12EI cos2L3+

6EI cosL2

-

Vertical Deformation

231

4 EIL

6 EI sin L2

6 EI cos L2

2 EIL

-

6 EI sin L2

6 EI cos L2

4 EIL

Rotational Deformation

232

( )cos + ( )sin

bxax

- byay

-AEL

N =

a

b

bx

by

ax

ay

[ ]

Internal forces

Normal Force

233

Internal forces

Bending moment

MAB= ( 2 + ) M(FER) AB +2 EI

L3 L-

MBA=2 EI

L( + 2 )

M(FER) BA +3 L-

234

Questions

235

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