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A. POMO CNI PROGRAMI

A.4 New in jlinpro - internal elastic links

Internal links are implemented so that we can solve problemssuch as structure on figureA.19, that has springs in corners.

p = 10kN/m

1.5m 2.5m 3.0m

3.0m

2.0m

cm = 12.000kNm

cm = 12.000kNm

EI = 50.000kNm2

EA =

Slika A.19

In joint with spring we define two nodes, (2 and 7 left, and 5and8 in the right joint), see figureA.20. These two nodes inone joint, have the same translational degrees of freedom anddifferent rotational degrees of freedom. Rotational degrees offreedom will be coupled with link elements, which have stiff-ness matrix

K =

k k

k k

1

0

2

4

3

5

4

3

6

7

8

9

7

8

10

11

12

13

11

12

14

15

16

17

18

19

20

18

19

21

22

23

24

0

1

2

7 4 83

9

10

5

6

6

Slika A.20: Numbering nodes and degrees of freedom

Listing A.7: Definition of degrees of freedom

defdofs

0,0,1,2

1,3,4,5

2,7,8,9

3,11,12,134,15,16,17

5,18,19,21

6,22,23,24

7,7,8,10

8,18,19,20

9,3,4,6

10,11,12,14

enddefdofs

Degrees of freedom are defined with pair of defdofs andenddefdofs, and in between we put node number, dof1, dof2,dof3. For given structure dof definition is given in listing A.7

Complete input file is given in file Framewithsprings.lscr

Listing A.8: Complete input file, FramewithSprings.lscr

#number of dofs per node

dimension,3

#nodes

n,0,0

n,0,3

n,0,5

n,1.5,7

n,4,7

n,7,5

n,7,0

n,0,7

n,7,7

n,1,2

n,2.5,4

# dof definition

# in case of manuall dof definition

# dofs have to be defined for each node

defdofs

0,0,1,2

1,3,4,5

2,7,8,9

3,11,12,13

4,15,16,17

5,18,19,21

6,22,23,24

7,7,8,10

8,18,19,20

9,3,4,6

10,11,12,14

enddefdofs

#cross sections

cs,50000,1E6,1

#elements

beam2D,0,1

beam2D,1,2

beam2D,7,3

beam2D,3,4,release,mj

beam2D,4,8

beam2D,5,6

beam2D,9,10

#links

#link,nodei,nodej,localdof,stiffnesslink,2,7,2,12000

link,5,8,2,12000

#supports

#spring,node,localdof,stiffness

spring,0,0,1E10

spring,0,1,1E10

spring,6,0,1E10

spring,6,1,1E10

#loads

py,-10

#load application

al,0,2,3,4

Run calculation

Now since for calculation of beam element stiffness matrix onlylength is calculated from position of nodes, and lengths are cor-rect compared to original structure,and since link stiffness ma-

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A.4 NEW IN JLINPRO - INTERNAL ELASTIC LINKS

trix is not calculated using any length but stiffness is inserteddirectly,and since global structural stiffness matrix is assem-bled based on DOF numbers that we defined manually, we ex-pect jlinpro to be able to solve such splitstructure correctly, sopush buttonRun.

Slika A.21: Moment diagram

Slika A.22: Axial force diagram

Now we will calculate this structure by the Force method inorder to compare results.

X1

X1

p = 10kN/m

1.5m 2.5m 3.0m

3.0m

2.0m

cm = 12.000kNm

cm = 12

.000

kNm

Slika A.23: Primary system

Rjesenje Primary system is shown on figureA.23. Reactions

A : VB 7 10 7 3.5 = 0

VB = 35kN

B : VA 7 10 7 3.5 = 0

VA= 35kN

CD : VB 3 10 3 1.5 HB 5 = 0

HB = 12kN

CL : VA 4 10 4 2 HA 5 = 0

HA= 12kN

p = 10kN/m

A B

C

4m 3.0m

3.0m

2.0m

HA HB

VA VB

Slika A.24

Moment diagram from external load is shown on figureA.25and from unit force on figureA.26

Calculation of flexibility coefficients

In order to calculate10 we have to calculate integral

l

M0M1

EI

that in range from 2.5 to 4 is not given in tables, (see figureA.25)so we will calculate it without tables. Coordinate systemis placed in hinge with x-axis pointing left, because it is easierto write moment function. Shear force in hinge is 5kN.

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A. POMO CNI PROGRAMI

36

60

60

60

60

M0

-

- -

-x2.54

Slika A.25: Bending moment diagram from external loading.

Bending moment function is

M0(x) = 5 x 5 x2

We calculate area and centroid position.

AM0 = 4

2.5

(5 x 5 x2) dx = 56.25

xA

T =

4

2.5

x (5 x 5 x2) dx

A =

190.547

56.25 = 3.3875

finally

1

EI

4

2.5

M0M1 = 1

EIAM0 M1(x

A

T) =

1

EI56.25

1.2

1.5 (3.3875 2.5) =

1

EI39.938

The rest of calculation of flexibility coefficient can be calculatedusing tables

EI10 = 2 1.26

(36 + 2 60) + EI 60 1.2cM

+ 39.938 vec prorac.

= 62.4 + 300 + 39.938 = 402.338

1.2

1.

2

M1

-

-

1.0

1.0

Slika A.26: Bending moment from unit force

Coefficient 11 is calculated using tables

EI 11 = 1

3 2 1.2 1.2

+1

3 1.5 1.2 1.2

+ EI MC1 MC1

cm= 7.68

Finally we obtain unknown force

X1 = 402.338

7.68 = 52.388kN

-36

2.87 2.

87 -

18.

75

-60

-60

M

-

- -

-

Slika A.27: Bending moment diagram

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