Stiffness 9

Lecture No. : 9 ال تاسعةالمحاضرة

F = K Dl l l

T K =g K l T

T

m m

Drive the member local stiffness matrix

Obtain the member global stiffness matrix

Drive the member transformation matrixT

Solution Steps of assembly method :Remember

Make assembly F = K D

Kuu

Kru

Kur

Krr

Fu

Fr

Du

Dr=

Make partition

Kgm K

gm K

gm K

gm

Remember

Kuu

Kru

Kur

Krr

Fu

Fr

Du

Dr=

Extract the stiffness equation

KuuFu Du= Kur Dr+

KuuDu =-1{ }Fu Kur Dr-

Obtain the deformation

Remember

Find internal forces in members

Calculate the reactions

KruFr Du= Krr Dr+

gF = K l lm m mT D

T

Remember

d1

d2

d3

d1

d2

Normal Force doesn’t taken

Drive the member local stiffness matrix

k11F1

F2

=k21

F3 k31

k12

k22

k32

k13

k23

k33

F4k41 k42 k43

k14

k24

k34

k44

d1

d2

d3

d4

d1

d2 d4

d3

First column inLocal Stiffness matrix

d1 =1

6 EI

L2

6 EIL2

12 EIL3

12 EIL3

d3

d4

d2

d2

12 EIL3

F2 =

F3 = -

F4 =

F1 =12 EI

L3

6 EIL3

6 EIL3

6 EI

L2

6 EIL2

12 EIL3

12 EIL3

=k31

12 EIL3

6 EIL2

-12EIL3

6 EIL2

k11

k21

k41

Second column inLocal Stiffness matrix

d2 =1

4 EIL

2 EIL

6 EIL2

6 EIL2

d3

d4

d1

d2

F3 =F1 =

F4 =F2 =

6 EIL2

6 EIL2

-

2 EIL

4 EIL

4 EIL

2 EIL

6 EIL2

6 EIL2

Second column inLocal Stiffness matrix

=

k12

k22

k32

k42

6 EI

L2

4 EIL

6 EIL2-

2 EIL

Third column inLocal Stiffness matrix

d3 =1

6 EIL2

6 EIL2

12 EIL3

12 EIL3

d3

d4

d1

d2

12 EIL3

6 EIL2

6 EIL2

12 EIL3F1 = F3 =

F4 =F2 = - -

-

6 EI

L2

6 EIL2

12 EIL3

12 EIL3

Third column in Local Stiffness matrix

=

k13

k23

k33

k33

12 EIL3

6 EIL2

12 EIL3

6 EIL2

-

-

-

Fourth column inLocal Stiffness matrix

d4 =1

4 EIL

2 EIL

6 EIL2

6 EIL2

d3

d4

d1

d2

F1 = F3=

F2 = F4=

6 EIL2

2 EIL

4 EIL

6 EIL2

-

4 EIL

2 EIL

6 EIL2

6 EIL2

Fourth column inLocal Stiffness matrix

=

k14

k24

k34

k44

6 EI

L2

2 EIL

6 EIL2-

4 EIL

K l

12 EIL3

6 EIL2

6 EIL2

4 EIL

-12 EIL3

=-12 EI

L3

-6 EIL2

-6 EIL2

12 EIL3

-6 EIL2

6 EIL2

2 EIL

6 EIL2

2 EIL

-6 EIL2

4 EIL

K l

12 EIL3

6 EI

L2

6 EI

L2

4 EIL

-12 EIL3

=-12 EI

L3

-6 EIL2

-6 EIL2

12 EIL3

-6 EIL2

6 EI

L2

2 EIL

6 EI

L2

2 EIL

-6 EIL2

4 EIL

K l = EIL3

K l

12 EIL3

6 EIL2

6 EIL2

4 EIL

-12 EIL3

= -12 EIL3

-6 EIL2

-6 EIL2

12 EIL3

-6 EIL2

6 EIL2

2 EIL

6 EIL2

2 EIL

-6 EIL2

4 EIL

12

6 L

-12

6 L

6 L

-6 L

4 L2

2 L2

-12

6 L

12

6 L

-6 L

2 L2

4 L2

6 L

d1

d2

Normal Force doesn’t taken

d1

If Shear is omitted

Drive the member local stiffness matrix

k11F1

F2

=k21

k12

k22

d1

d2

d1 d2

K l

12 EIL3

6 EIL2

6 EIL2

4 EIL

-12 EIL3

=-12 EI

L3

-6 EIL2

-6 EIL2

12 EIL3

-6 EIL2

6 EIL2

2 EIL

6 EIL2

2 EIL

-6 EIL2

4 EIL

K l4 EI

L=

2 EIL

2 EIL

4 EIL

K l = 2EIL 1

2

2

1

Construct the stiffness matrix for the shown beam where EI is

constant for all members

Example 1:

8 10

A CB

First element : (A-B )Start Joint : A End Joint : B

K l = EIL3

12

6 L

-12

6 L

6 L

-6 L

4 L2

2 L2

-12

6 L

12

6 L

-6 L

2 L2

4 L2

6 L

K l K g=

K l

12 EIL3

6 EI

L2

6 EI

L2

4 EIL

-12 EIL3

=-12 EI

L3

-6 EIL2

-6 EIL2

12 EIL3

-6 EIL2

6 EI

L2

2 EIL

6 EI

L2

2 EIL

-6 EIL2

4 EIL

K l = EIL3

12

6 L

-12

6 L

6 L

-6 L

4 L2

2 L2

-12

6 L

12

6 L

-6 L

2 L2

4 L2

6 L

K l = EI

.023

.0937

.-023

.0937

.0937

.5

.0937

.25

.-023

.0937

.023

.-0937

.0937

.25

.-0937

.5

A

B

�ِA

B

Second element : ( B-c)Start Joint : B End Joint : c

K l

12 EIL3

6 EI

L2

6 EI

L2

4 EIL

-12 EIL3

=-12 EI

L3

-6 EIL2

-6 EIL2

12 EIL3

-6 EIL2

6 EI

L2

2 EIL

6 EI

L2

2 EIL

-6 EIL2

4 EIL

K l = EI

.012

.06

.-012

.06

.06

.4

.-06

.2

.-012

.-06

.012

.-0937

.06

.20

.-06

.4

B C

C

B

Assembly :

K =g1 EI

.023

.0937

.-023

.0937

.0937

.5

.0937

.25

.-023

.0937

.023

.-0937

.0937

.25

.-0937

.5

A B

B

A

EI .06

.-012

.06

.06

.4

.2

.-012

.-06

.012

.-06

.06

.2

.-06

.012

.-06

K =g2

.4

B C

B

C

Ks = EI

.023

.0937

.-023

.093700

.0937

.5

.0937

.2500

.-023

.0937

.035

.-0337.-012.06

.0937

.25

.-0337

.9

.-06

.2

0

0

.-012

.-06

.012.-06

0

0

.06

.2

.-06

.4

A B C

B

A

C

Partition

KuuK =Kru

Kur

Krr

u ru

r

Kuu = EI .90

Example 2:Construct the stiffness matrix for the shown beam where EI is constant for all members

4254

ECB DA

= E I

12/L^3

6/L^2

-12/L^3

6/L^2

6/L^2

2/L

6/L^2-

4/L

12/L^3-

6/L^2-

12/L^3

6/L^2-

6/L^2

-6/L^2

4/L

2/L

K gK l =

K l

First element : (A-B )Start Joint : A

End Joint : BAngle : 0s = sin = 0c = cos = 1

Is conastant EA

LAB = 400 cm

Kl= EI

.1875

.375

.-1875

.375

.375

1

.-375

.5

.-1875

.-375

.1875

.-375

.375

.5

.-375

1

A B

B

A

Second element : B-C( Start Joint : B

End Joint : cAngle : 0s = sin = 0c = cos = 1

Is conastant EA

LBC = 500 cm

EI .24

.-096

.24

.24

.8

.4

.-096

.-24

.096

.-24

.24

.4

.-24

.096

.-24

.8

B C

B

C

K =g1

End Joint : D

Third element : (C-D )

Start Joint : C

Angle : 0s = sin = 0c = cos = 1

LCD = 200 cm

Is conastant EA

= EI

1.5

1.5

-1.5

1.5

1.5

2

-1.5

.2

-1.5

-1.5

1.5

-1.5

1.5

1

-1.5

.2

C D

D

CK l

Fourth element : (D-E )

Start Joint :D

End Joint : EAngle : 0s = sin = 0c = cos = 1

Is conastant EA

LD-E = 400 cm

Kl= EI

.1875

.375

.-1875

.375

.375

1

.-375

.5

.-1875

.-375

.1875

.-375

.375

.5

.-375

1

D "ُE

D

E

K =s

.1875 .375.375 1

.-1875 .-375

.375 .5

0 0

0 0

0 0

0 0

.-096 .-24.24 .4

0 00 0

0

0

0 00

00

0 0

0 0

0

0

0

0

0

0 00

0.-1875 .375.-375 .5

.2835 .-135.-135 1.8

0000

.-096

.-24

.24.4

1.596 1.261.26 2.8

-1.5 1.51.5 1

0 00 0

00

-1.5 1.51.5 1

1.687 -1.125-1.125 3.-1875.375

.-375.5

0

00

0

0

0

.-1875 .375.-375 .5

.1875.-375 1

.-375

A B C D E

AB

CDE

KUU1.8 EI0.4 EI

0

0.4 EI2.8 EI

EI

0EI

3 EI=

1.80.4

0

0.42.8

1

01

3EI=

Example 3:Calculate the deformation of the shown beam where EI = 105 kN.m2 for all members

4254DCB

EA 30 kNm50 kNm60 kNm

KUU

1.8 EI0.4 EI

0

0.4 EI2.8 EI

EI

0EI

3 EI=

1.80.4

0

0.42.8

1

01

3EI

The stiffness equation

F = K D

Stiffness matrix From Exampl (2)

F1

F2

=-60

-50

F3 30

= d1

d2

-60

-50

1.8

.4

.4

2.8

F = K D

30 d3

0

1

0 1 3

D =

B

C

D

D = K-1 F- 60- 50

30=

1.80.4

0

0.42.8

1

01

3

B

C

D

1EI

-1

- 60- 50

30=

7.4-1.2

.4

-1.25.4

-1.8

.4-1.8

4.88

B

C

D

1105

112.84

=

- 0.290- 0.196

0.165

X10-3

rad

Example 4:Draw B.M.D for the shown beam where EI = 105 kN.m2 for all members

4

A

B

E

5 2 4C D

60 kNm 50 kNm 30 kNm

From the previous example

=

B

C

D

- 0.290

- 0.196

0.165

X10-3

rad

4

AB

E

5 2 4C D

=

B

C

D

- 0.290- 0.196

0.165

X10-3

rad

EI = 105 kN.m2

For member AB

LAB

MAB = == - 14.5 kN.m

2 EILAB

MBA= (2 B + A ) = 5x104(2X-0.029)x10-3

5x104(0-0.29)x10-3(2 A + B )2 EI

= - 29 kN.m

4

AB

E

5 2 4C D

=C

D

- 0.290- 0.196

0.165

X10-3

rad

B

EI = 105 kN.m2

For member BC

LBC

MBC = (2 B + C ) =

2 EI

LBC

MCB= (2 C + B ) =

4x104(2x-0.29-0.196)x10-3

= - 31 kN.m

4x104(2x-0.196-0.029)x10-3

= - 27.3 kN.m

MAB = - 14.5 kN.mMBA= - 29 kN.m

MBC = - 31 kN.mMCB= - 27.3 kN.m

MCD = - 22.7 kN.mMDC= 13.5 kN.m

MDE = 16.5 kN.mMED= 8.3 kN.m

A B B C C D D E

14.5 29 31 27.3 22.7 13.5 16.5 8.3

A B B C C D D E

14.5 29 31 27.3 22.7 13.5 16.5 8.3

14.531

27.3

22.713.5

16.5

B.M.D

14.5 13.5

16.527.3

22.731

29

A

B

E

C D

60 kNm 50 kNm 30 kNm

Example 5:Draw B.M.D for the shown beam where EI is constant for all members

2

A B C D

1 2 2 1 2

100 kN 200 kN 150 kN

Solution Steps of assembly method :

Drive the member local stiffness matrixLocal

k11F1

F2

=k21

F3 k31

k12

k22

k32

k13

k23

k33

F4k41 k42 k43

k14

k24

k34

k44

d1

d2

d3

d4

d3

d1

d2 d4

First column inLocal Stiffness matrix

6 EI

L2

6 EI

L212 EI

L3

F1 = 12 EIL3

F2 = 6 EI

L2

6 EI

L2

F3 = 12 EIL3

-

F4 =

first column in Local Stiffness matrix

=

k31

k41

12 EIL3

6 EIL2

-12 EIL3

6 EIL2

k11

k21

Second column inLocal Stiffness matrix

d2 =1

4 EIL

2 EIL

6 EIL2

6 EIL2

F3 =F1 =

F4 =F2 =

6 EIL2

6 EIL2

-

2 EIL

4 EIL

Second column inLocal Stiffness matrix

=

k12

k22

k32

k42

6 EI

L2

4 EIL

6 EIL2

- 2 EIL

Third column in Local Stiffness matrix

d3 =1

12 EIL3

12 EIL3

6 EIL2

6 EIL2

Third column in Local Stiffness matrix

=

k13

k23

k33

k33

-12 EIL3

-6 EIL2

12 EIL3

-6 EIL2

12 EIL3

-6 EIL2

-6 EIL2

F4 =

F3 = F1 = -12 EIL3

F2 =

Fourth column inLocal Stiffness matrix

6 EIL2

6 EIL2

4 EIL

2 EIL

F1 = F3=

F2 = F4 =

6 EIL2

2 EIL

-6 EIL2

4 EIL

K l

12 EIL3

6 EIL2

6 EIL2

4 EIL

-12 EIL3

=

-12 EIL3

-6 EIL2

-6 EIL2

12 EIL3

-6 EIL2

6 EIL2

2 EIL

6 EIL2

2 EIL

-6 EIL2

4 EIL

First element : (A-B )

Start Joint : A

End Joint : BAngle : 0s = sin = 0c = cos = 1EA Is conastant

LAB = 300 cm

K l K g=

K l = E I

12/L^3

6/L^2

-12/L^3

6/L^2

6/L^2

2/L

6/L^2-

4/L

12/L^3-

6/L^2-

12/L^3

6/L^2-

6/L^2

-6/L^2

4/L

2/L

K l = EI

0.44

0.67

-0.44

0.67

0.67

1.33

-0.67

0.67

-0.44

-0.67

0.44

-0.67

0.67

0.67

-0.67

1.33

A B

"ِ"ِA

B

Second element : ( B-c)Start Joint : B End Joint : c

= E I

12/L^3

6/L^2

-12/L^3

6/L^2

6/L^2

2/L

6/L^2-

4/L

12/L^3-

6/L^2-

12/L^3

6/L^2-

6/L^2

-6/L^2

4/L

2/LK l

K l = EI

0.1875

0.375

-0.1875

0.375

0.375

1

-0.375

0.5

-0.1875

-0.375

0.1875

-0.375

0.375

0.5

-0.375

1

B C

C

B

Second element : ( C-D )Start Joint : C End Joint : D

= E I

12/L^3

6/L^2

-12/L^3

6/L^2

6/L^2

2/L

6/L^2-

4/L

12/L^3-

6/L^2-

12/L^3

6/L^2-

6/L^2

-6/L^2

4/L

2/LK l

K l = EI

0.44

0.67

-0.44

0.67

0.67

1.33

-0.67

0.67

-0.44

-0.67

0.44

-0.67

0.67

0.67

-0.67

1.33

C D

�ِC

D

K

g = EI

0.44

0.67

-0.44

0.67

0.67

1.33

-0.67

0.67

-0.44

-0.67

0.44

-0.67

0.67

0.67

-0.67

1.33

A B

�Aِ

B

K g= EI

0.1875

0.375

-0.1875

0.375

0.375

1

-0.375

0.5

-0.1875

-0.375

0.1875

-0.375

0.375

0.5

-0.375

1

B C

C

B

K g= EI

0.44

0.67

-0.44

0.67

0.67

1.33

-0.67

0.67

-0.44

-0.67

0.44

-0.67

0.67

0.67

-0.67

1.33

C D

�ِC

D

Ks = EI

0.44

.0937

0.67

-0.44

0.67

0

0

0.67

1.33

-0.67

0.67

0

0

0.67

0.67

-.0.295

2.33

-0.375

0.5

0

0

-0.1875

.-375

.2525

0.295

0

0

0.375

0.5

0.295

2.33

0

0

.-44

.-67

A B C

B

A

C

K l K g=

00 0 .-44 .-67 .-67

0 0 0 .67 0.67 1.33

D

D

-0.44

-0.67

.0.6275

-0.295

-0.1875

0.375

0

0

0

0

.67

0.67

0

0

0

0

.-67

0.44

Partition

KuuK =Kru

Kur

Krr

u ru

r

Kuu = EI 2.33

2.33

0.5

0.5

Force vectorTransformation from member forces to Joint forces

L

P

8LP

8LP

90

P a b2

L2

L

P

a b P b a2

L2

2

A B C D

1 2 2 1 2

100 kN 200 kN 150 kN

100 kN 150 kN

200 kN

100 kNm100 kNm

44.4 kNm

22.2 kNm

66.7 kNm

33.3 kNm

Fixed End Reaction

(FER)

2

A B C D

1 2 2 1 2

100 kN 200 kN 150 kN

100 kN 150 kN

200 kN

100 kNm100 kNm

44.4 kNm

22.2 kNm

66.7 kNm

33.3 kNm

Fixed End Action (FEA)

2

A B C D

1 2 2 1 2

100 kN 200 kN 150 kN

100 kNm

44.4 kNm 66.7 kNm

100 kNm

33.3 kNm55.6 kNm

2

A B C D

1 2 2 1 2

100 kN 200 kN 150 kN

33.3 kNm55.6 kNm

F1

F2

=-55.6

33.3

F = K Dk11F1

F2

=k21

k12

k22

d1

d2

-55.6

33.3=

7/3

0.5 7/3

0.5EI

B

C

B

C=

1EI

7/3

0.5 7/3

0.5-1

-55.6

33.3

B

C=

1EI

7/3

0.5 7/3

0.5-1

-55.6

33.3

B

C=

1EI

-28.18

20.31

Internal forces in beam elements2 EI

LMAB= ( 2 + )

MBA=2 EI

L( + 2 )

MBA=2 EI

L( + 2 ) M(FER) BA +

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

L

2

A B CD

1 2 2 1 2

100 kN 200 kN 150 kN

100 kN 150 kN

200 kN

100 kNm100 kNm

44.4 kNm

22.2 kNm

66.7 kNm

33.3 kNm

Fixed End Reaction

(FER)

B

C=

1EI

-28.18

20.31

2 EILMAB= ( 2 + ) M(FER) AB +

MBA=2 EI

L( + 2 ) M(FER) BA +

100 kN

44.4 kNm22.2 kNm

= 22.2 + 2/3 (-28.18) = 3.4

= -44.4 + 2/3 (2x-28.18) = - 82

A B

B

C=

1EI

-28.18

20.31

2 EILMBC= ( 2 + ) CM(FER) BC +

MCB=2 EI

L( + 2 ) CM(FER) CB +

= 100 + 2/4 (2x-28.18+20.31) = 82

= -100 + 2/4 (-28.18+2x20.31) = - 93.8

200 kN

100 kNm100 kNm

B C

B

C=

1EI

-28.18

20.31

2 EILMCD= ( 2 + )C DM(FER) CD +

MDC=2 EI

L( + 2 )C DM(FER) DC +

= 66.7 + 2/3 (2x20.31) = 93.8

= -33.3 + 2/3 (20.31) = - 19.8

150 kN

33.3 kNm66.7 kNm

C D

MAB= 3.4MBA= -82

MBC= 82MCB= -93.8

MCD= 93.8MDC= -19.8

3.4

82

19.8

93.8

B.M.D

3.4

82

19.8

93.8

B.M.D

2

A B C D

1 2 2 1 2

100 kN 200 kN 150 kN

55.887.9

69.1

66.7

10.9

200

112.1

100

30.9

Example 6:Draw B.M.D for the shown beam where EI is shown in figure

3

A B C D

5 5 2

100 kN 200 kN 100 kN

3 2

250 kN

III2 I2 I2 I

E

Solution Steps of assembly method :

Drive the member local stiffness matrixLocal

k11F1

F2

=k21

F3 k31

k12

k22

k32

k13

k23

k33

F4k41 k42 k43

k14

k24

k34

k44

d1

d2

d3

d4

d3

d1

d2 d4

First column inLocal Stiffness matrix

6 EI

L2

6 EI

L212 EI

L3

F1 = 12 EIL3

F2 = 6 EI

L2

6 EI

L2

F3 = 12 EIL3

-

F4 =

first column in Local Stiffness matrix

=

k31

k41

12 EIL3

6 EIL2

-12 EIL3

6 EIL2

k11

k21

Second column inLocal Stiffness matrix

d2 =1

4 EIL

2 EIL

6 EIL2

6 EIL2

F3 =F1 =

F4 =F2 =

6 EIL2

6 EIL2

-

2 EIL

4 EIL

Second column inLocal Stiffness matrix

=

k12

k22

k32

k42

6 EI

L2

4 EIL

6 EIL2

- 2 EIL

Third column in Local Stiffness matrix

d3 =1

12 EIL3

12 EIL3

6 EIL2

6 EIL2

Third column in Local Stiffness matrix

=

k13

k23

k33

k33

-12 EIL3

-6 EIL2

12 EIL3

-6 EIL2

12 EIL3

-6 EIL2

-6 EIL2

F4 =

F3 = F1 = -12 EIL3

F2 =

Fourth column inLocal Stiffness matrix

6 EIL2

6 EIL2

4 EIL

2 EIL

F1 = F3=

F2 = F4 =

6 EIL2

2 EIL

-6 EIL2

4 EIL

K l

12 EIL3

6 EIL2

6 EIL2

4 EIL

-12 EIL3

=

-12 EIL3

-6 EIL2

-6 EIL2

12 EIL3

-6 EIL2

6 EIL2

2 EIL

6 EIL2

2 EIL

-6 EIL2

4 EIL

First element : (B-C )

Start Joint : B

End Joint : CAngle : 0s = sin = 0c = cos = 1EA Is conastant

LAB = 1000 cm

K l K g=

K l =E I

12/L^3

6/L^2

-12/L^3

6/L^2

6/L^2

2/L

6/L^2-

4/L

12/L^3-

6/L^2-

12/L^3

6/L^2-

6/L^2

-6/L^2

4/L

2/L

K l = EI

0.024

0.12

-0.024

0.12

0.12

0.8

-0.12

0.4

-0.024

-0.12

0.024

-0.12

0.12

0.4

-0.12

0.8

B C

�ِB

C

First element : (C-D )

Start Joint : C

End Joint : DAngle : 0s = sin = 0c = cos = 1EA Is conastant

LAB = 500 cm

K l K g=

K l = E I

12/L^3

6/L^2

-12/L^3

6/L^2

6/L^2

2/L

6/L^2-

4/L

12/L^3-

6/L^2-

12/L^3

6/L^2-

6/L^2

-6/L^2

4/L

2/L

K l = EI

0.096

0.24

-0.096

0.24

0.24

0.8

-0.24

0.4

-0.096

-0.24

0.096

-0.24

0.24

0.4

-024

0.8

C D

C

D

Assembly :

K g= EI

0.024

0.12

-0.024

0.12

0.12

0.8

-0.12

0.4

-0.024

-0.12

0.024

-0.12

0.12

0.4

-0.12

0.8

B C

�ِB

C

K g= EI

0.096

0.24

-0.096

0.24

0.24

0.8

-0.24

0.4

-0.096

-0.24

0.096

-0.24

0.24

0.4

-024

0.8

C D

C

D

Ks = EI

0.024

.0937

0.12

-0.024

0.12

0

0

0.12

0.8

-0.12

0.4

0

0

-0.024

-0.12

0.12

0.12

-0.096

0.024

0.12

0.4

0.12

1.6

-0.024

0.4

0

0

-0.096

-0.24

0.096

-0.24

0

0

0.24

0.4

-0.24

0.8

B C D

C

B

D

K l K g=

Partition

KuuK =Kru

Kur

Krr

u ru

r

Kuu = EI 0.8

0.4

0

0.4

1.60.4 0.8

0.4

0

Force vectorTransformation from member forces to Joint forces

L

P

8LP

8LP

128

P a b2

L2

L

P

a b P b a2

L2

100 kN

100 kN200 kN

250 kNm250 kNm

300 kNm

200 kNm

Fixed End Reaction (FER)

3

A B C D

5 5 2

100 kN 200 kN 100 kN

3 2

250 kN

E

250 kN180 kNm 120 kNm

100 kN

100 kN200 kN

250 kNm250 kNm

300 kNm

200 kNm

Fixed End Action (FEA)

3

A B C D

5 5 2

100 kN 200 kN 100 kN

3 2

250 kN

E

250 kN180 kNm 120 kNm

250 kNm250 kNm

300 kNm

200 kNm

3

A B C D

5 5 2

100 kN 200 kN 100 kN

3 2

250 kN

E

180 kNm 120 kNm

50 kNm 70 kNm 80 kNm

3

A B C D

5 5 2

100 kN 200 kN 100 kN

3 2

250 kN

E

50 kNm 70 kNm 80 kNm

F1

F2 =

50

70

F3 - 80

F = K DThe stiffness equation

=

50

70- 80

0.80.4

0

0.41.6

0.4

00.4

0.8EI

B

C

D

=

50

70- 80

0.80.4

0

0.41.6

0.4

00.4

0.8

B

C

D

1EI

-1

=1EI

27.08

70.83- 135.42

=

B

C

D

1EI

27.08

70.83- 135.42

Internal forces in beam elements

2 EILMAB= ( 2 + )

MBA=2 EI

L( + 2 )

M(FER) AB +

M(FER) BA +

100 kN

100 kN200 kN

250 kNm250 kNm

300 kNm

200 kNm

Fixed End Reaction (FER)

3

A B C D

5 5 2

100 kN 200 kN 100 kN

3 2

250 kN

E

250 kN180 kNm 120 kNm

2 E(2I)LMBC= ( 2 + ) CM(FER) BC +

MCB=2 E(2I)

L( + 2 ) CM(FER) CB +

= 250 + 4/10 (2x27.08+70.83) = 300

= -250 + 4/10 (27.08+2x70.83) = - 182.5

200 kN

250 kNm250 kNm

B C

=

B

C

D

1EI

27.08

70.83- 135.42

2 E(2I)LMBC= ( 2 + ) CM(FER) BC +

MCB=2 E(2I)

L( + 2 ) CM(FER) CB +

= 250 + 4/10 (2x27.08+70.83) = 300

= -250 + 4/10 (27.08+2x70.83) = - 182.5

200 kN

250 kNm250 kNm

B C

=

B

C

D

1EI

27.08

70.83- 135.42

MBC= 300MCB= -182.5

MCD= 182.5MDC= -200

300200

B.M.D

A

B C D

E

182.5

B.M.D

3

A B C D

5 5 2

100 kN 200 kN 100 kN

3 2

250 kN

E

300200

A

B C D

E

182.5241.25

500

258.75

189.5

300

110.5

Example 7:Draw B.M.D for the shown beam where EI is constant for all members

A B

5 5

240 kN

C

5 5

120 kN

K l

12 EIL3

6 EIL2

6 EIL2

4 EIL

-12 EIL3

=

-12 EIL3

-6 EIL2

-6 EIL2

12 EIL3

-6 EIL2

6 EIL2

2 EIL

6 EIL2

2 EIL

-6 EIL2

4 EIL

4 EIL

First element : (A-B )

EA conastant

LAB = 10 m

12 EIL3

=0.012EI

6 EIL2

4 EIL

2 EIL

=0.06 EI

=0.4 EI

=0.2 EI

K l =

0.012 EI

0.012 EI 0.012 EI

0.012 EI

0.06 EI

0.06 EI

0.06 EI0.06 EI

0.06 EI

0.06 EI

0.06 EI

0.06 EI

0.4 EI

0.4 EI

0.2 EI

0.2 EI

A

A

B

B

Second element : ( B-c)

EI conastant LAB = 10 m

12 EIL3

=0.012EI

6 EIL2

4 EIL

=0.06 EI

=0.4 EI

=0.2 EI2 EI

L

K l =

0.012 EI

0.012 EI 0.012 EI

0.012 EI

0.06 EI

0.06 EI

0.06 EI0.06 EI

0.06 EI

0.06 EI

0.06 EI

0.06 EI

0.4 EI

0.4 EI

0.2 EI

0.2 EI

A

A

B

B

K gK l =

Assembly :

=

K gg1

0.012 EI 0.06 EI 0.012 EI 0.06 EI

0.06 EI 0.4 EI 0.06 EI 0.2 EI

0.012 EI 0.06 EI 0.06 EI0.012 EI

0.06 EI 0.2 EI 0.4 EI0.06 EI

2=

K gg

0.012 EI

0.012 EI 0.012 EI

0.012 EI

0.06 EI

0.06 EI

0.06 EI0.06 EI

0.06 EI

0.06 EI

0.06 EI

0.06 EI

0.4 EI

0.4 EI

0.2 EI

0.2 EI

A B

A

B

=

0.012

-0.012

-0.012

0.06

0.06

-0.06 0.06

0.06

-0.06

0.06

-0.06

0.4

0.4

0.2

0.2

A

A

B

B

K s

0.0

0.8

0.024

0.0

0.0

0.0

0.0 0.0

0.0

0

0.0

0.0

0.012

-0.012

-0.012

0.06

-0.06

0.2

0.2

-0.06

-0.06

EI

c

c

Partition

KuuK =Kru

Kur

Krr

u ru

r

Force vectorTransformation from member forces to Joint forces

L

P

8LP

8LP-

240 kN300 kNm300 kNm

Fixed End Reaction

(FER)

A B

5 5

240 kNC

5 5

120 kN

120 kN150 kNm150 kNm

300 kNm300 kNm

A B

5 5

240 kNC

5 5

120 kN

150 kNm150 kNm

120120 6060

F1

F2

=

120

-300

F3

F6

F5

F4

180

150

60

150

120

-300

180

150

60

150

0.012

-0.012

-0.012

0.06

0.06

-0.06 0.06

0.06

-0.06

0.06

-0.06

0.4

0.4

0.2

0.2

0.0

0.8

0.024

0.0

0.0

0.0

0.0 0.0

0.0

0

0.0

0.0

0.012

-0.012

-0.012

0.06

-0.06

0.2

0.2

-0.06

-0.06

d1

d2

d3

d4

d5

d6

= EI

120

-300

180

150

60

150

-0.012 0.06

-0.06

0.06

-0.06

0.06

-0.06

0.4

0.4

0.2

0.2

0.0

0.8

0.024

0.0

0.0

0.0

0.0 0.0

0.0

0.0

0.0

0.0

0.012

-0.012

-0.012

0.06

-0.06

0.2

0.2

-0.06

-0.06

d1

d2

d3

d4

d5

d6

=EI

=B

C

150

150

0.8

0.2 0.4

0.2EI

B

C=

1EI

-1150

150

0.8

0.2 0.4

0.2

B

C=

1EI

107.14

321.43

B

C=

1EI

-1150

150

0.8

0.2 0.4

0.2

Internal forces in beam elements

2 EILMAB= ( 2 + )

MBA=2 EI

L( + 2 )

M(FER) AB +

M(FER) BA +

240 kN300 kNm300 kNm

Fixed End Reaction

(FER)

A B

5 5

240 kNC

5 5

120 kN

120 kN150 kNm150 kNm

2 EILMAB= ( 2 + )

MBA=2 EI

L( + 2 )

M(FER) AB +

M(FER) BA +

= 300 + 2/10 (107.14) = 321.4

= -300 + 2/10 (2x107.14) = - 257.1

B

C=

1EI

107.14

321.43

240 kN

300 kNm300 kNm

A B

2 EILMBC= ( 2 + ) C

MCB=2 EI

L( + 2 ) C

M(FER) BC +

M(FER) CB +

= 150 + 2/10 (2x107.14+321.43) = 257.1

= -150 + 2/10 (107.14+2x321.43) = 0

120 kN

150 kNm150 kNm

B C B

C=

1EI

107.14

321.43

MAB= 321.4MBA= -257.1

MBC= 257.1MCB= 0

321.4257.1

B.M.D

93.8

128.55

300

171.45

A B

5 5

240 kNC

5 5

120 kN

321.4257.1

B.M.D

289.25

600

310.75

Beams with settlement

6 EIL2

6 EIL2 12 EI

L3

12 EIL3

Fixed End Reaction

(FER)

Beams with settlement

6 EIL2

6 EIL2

12 EIL3

12 EIL3

Fixed End Reaction

(FER)

Beams with settlement

3 EIL2 3 EI

L3

3 EIL3

Fixed End Reaction

(FER)

Beams with settlement

3 EIL2

3 EIL3

3 EIL3

Fixed End Reaction

(FER)

169

Example 8:Draw B.M.D for the shown beam due to the shown loads and vertical downward settlement at support B (2000/EI) and at support C (1000/EI) where EI is constant for all members

A B

5 5

240 kN

C

5 5

120 kN

=0.8

0.2 0.4

0.2K EI

From example 7 :

240 kN300 kNm300 kNm

Fixed End Reaction

(FER)

A B

5 5

240 kNC

5 5

120 kN

120 kN150 kNm150 kNm

ForLoads

Fixed End Reaction (FER)

A B

5 5

C

5 5

Forsettlement

1000EI

2000EI

6 EIx1000102 EI

6 EIx1000102 EI

6 EIx2000102 EI

6 EIx2000102 EI

120

120 6060

300 kNm300 kNm

A B

5 5

240 kNC

5 5

120 kN

150 kNm150 kNmFor

Loads

Forsettlement 120 120 60 60

Fixed End Reaction

(FER)

210 kNmTotal 90 kNm

420 180 90 210

A B

5 5

240 kNC

5 5

120 kN

Fixed End Reaction

(FER)210 kNm90 kNm

Fixed End Action (FEA)

210 kNm90 kNm

A B

5 5

240 kNC

5 5

120 kN

Fixed End Action (FEA)

210 kNm90 kNm

F1

F2

=90

210

F = K Dk11F1

F2

=k21

k12

k22

d1

d2

=B

C

B

C=

1EI

-1

90

210

90

210

0.8

0.2 0.4

0.2EI

0.8

0.2 0.4

0.2

B

C=

1EI

-21.43

535.71

B

C=

1EI

-190

210

0.8

0.2 0.4

0.2

Internal forces in beam elements

2 EILMAB= ( 2 + )

MBA=2 EI

L( + 2 )

M(FER) AB +

M(FER) BA +

300 kNm300 kNm

A B

5 5

240 kNC

5 5

120 kN

150 kNm150 kNmFor

Loads

Forsettlement 120 120 60 60

Fixed End Reaction

(FER)

210 kNmTotal 90 kNm

420 180 90 210

2 EILMAB= ( 2 + )

MBA=2 EI

L( + 2 )

M(FER) AB +

M(FER) BA +

= 420 + 2/10 (-21.43) = 415.7

= -180 + 2/10 (2x-21.43) = - 188.6

180 kNm420 kNm

A B B

C=

1EI

-21.43

535.71

2 EILMBC= ( 2 + ) C

MCB=2 EI

L( + 2 ) C

M(FER) BC +

M(FER) CB +

= 90 + 2/10 (2x-21.43+535.7) = 188.6

= -210 + 2/10 (-21.43+2x535.71) = 0

210 kNm90 kNm

B C B

C=

1EI

-21.43

535.71

MAB= 321.4MBA= -257.1

MBC= 257.1MCB= 0

415.7188.6

B.M.D

93.8

94.3

300

205.7

A B

5 5

240 kNC

5 5

120 kN

415.7188.6

B.M.D

302.15

600

297.85