advanced structure

7/31/2019 Advanced Structure

1/20

University of Technology

Building & Construction Dept.

Structural Engineering Branch

Fourth Stage Structural Eng. Branch

Dr. Alaa K. Abdul Karim


2/20

Matrix Displacement (Stiffness) Method

:nIntroductio

The matrix displacement method, which is also known as the stiffness

method, is a method of analysis that uses the stiffness properties of the elements of

a structure to form a set of simultaneous equation relating displacements of the

structure to loads acting on the structure. The governing matrix equation is of the

form :

= .KF

Where F is a column matrix of external loads, K is the stiffness matrix of the

structure, and is a column matrix of external displacements. For a given set of

external loads, the corresponding displacements are obtained by solving the set of

simultaneous equations. Force in the elements of the structure are obtained through

the use of the calculated displacements and the element stiffness properties. In

analyzing a structure by the matrix displacement method, it is advantageous to

consider the structure in a broad fashion. A structure is considered to be an

assembly of structural elements connected at a finite number of points referred to

as nodal point and loaded only at these points.

The following basic conditions are satisfied at each node :

1. The equations of equilibrium.2. The compatibility of displacements.3. The force - displacement relationship.


3/20

In skeletal structures, it is assumed that the element of the structure is straight and

prismatic thus, the position of the nodes is located at places where abrupt changes

in geometry, loading or material properties occur. Examples of these natural

subdivisions are shown in Fig. below.

One Dimensional Elements :

A one dimensional element may be represented by a straight line whose ends,

such as 1 and 2, are nodal points. These elements are referenced in a coordinate

system Known as local or element coordinate system. In this system , x- axis is

defined by element axis which is a line joining the two nodes of the elements. One

- dimensional elements are used when the geometry, material properties, and

dependent variables such as displacements can all be expressed in terms of one

independent space coordinate which is measured along the element axis.


4/20

Pin-Jointed Bar Element:

The pin-jointed bar or truss element shown in Fig. below is the simplest

structural element and is assumed to be pin connected at both the ends. The bar

element is also assumed to have constant cross-sectional area (A) and modulus of

elasticity (E) over its length (L), external loads are applied at the nodes and effect

of self-weight is neglected. Thus for a plane structure, this element has four

degrees of freedom, two at each node.

Beam Element:

The beam element shown in Fig. below, is also known as frame element. For a

plane structure, this element has six degrees of freedom, three at each node, ( axial

and in-plane transverse displacements, and in-plane rotation.

Global axis

YLocal axis

Y'

F1

F2

F4

F3

X'

X

1

2

DOF= 4

Y

F1

F2

F5

F4

X1

2

DOF= 6

F3 F6


5/20

Stiffness Matrix of Elastic Spring

A typical elastic spring is shown in Fig. below. Forces exist at nodes 1 and 2.

Assume that node 2 is fixed and node 1 is displaced by 1 due to forceF1,1 (case A).

Appling Equilibrium of forces ( Fx = 0 )F1,1= - F2,1 = k1

Similarly assume that node 1 is fixed and node 2 is displaced by 2 due

to force F2,2 (case B).

Appling Equilibrium of forces ( Fx = 0 )F2,2= - F1,2 = k2

By using the principle of superposition for case A and B (adding up

algebraically the two cases A &B.) we obtain :

Node 1 Node 2

F1

, 1

F2 ,

2K

Node 1 Node 2

K1

F1,1 F2,1

Node 1 Node 2

K 2

F1,2 F2,2

Node 1 Node 2

K 2

F1 F2

1


6/20

F1= F1,1 + F 1,2

F 2= F 2,1 + F 2,2

Sub. The value of F1,1 , F 1,2 , F 2,1 and F 2,2

F 1= k1 - k2

F 2= - k1 + k2

In matrix notation, these equations may be re- written as :

=

2

1

2

1

.KK

kK

F

F

{ } [ ]{ }= .KF

Spring Assemblage

Actual structures generally consist of basic structural componentsstringers, beams, thin plate, etc. properly fastened into an assemblage.

For example, a truss is usually considered as an assemblage of axial

force members, whereas a building frame consists of an assemblage of

beams and columns. Consequently, it is vitally important to be able to

form the total structure stiffness matrices for the separate components.

Consider the assemblage of springs A and B shown:

Node 1 Node 2

Ka2

F1

F21

Node 3

Kb3

F3


7/20

The first step is to assemble the stiffness matrix for each element :

For spring a :

=

2

1

2

1.

aa

aa

KK

kK

F

F

For spring b :

=

3

2

3

2.

bb

bb

kK

kK

F

F

using the principle of superposition and applying the rule of matrix

addition, we obtain :

+

=

3

2

1

3

2

1

.

0

0

bb

bbaa

aa

KK

KKKK

KK

F

F

F

Example 1:Analyze the linear spring system shown in Fig. below which has two

linear springs connected a series with spring stiffness Ka and Kb .

Node 1 Node 2

KaF1

F2

Node 3

KbF3

Node 1 Node 2

Ka

Node 3

KbP

Element 1 Element 2


8/20

For element 1 : For element 2 :

=

2

1

2

1.

aa

aa

KK

kK

F

F

=

3

2

3

2.

bb

bb

kK

kK

F

F

overall stiffness matrix for all system (assemblage of system).

+

=

3

2

1

3

2

1

.

0

0

bb

bbaa

aa

KK

KKKK

KK

F

F

F

1 = 0F2 = 0

F3 = P

+=

3

2.

0

bb

bba

KK

kKK

P

solve the two equations to obtain the values of2 and 3

2 = P/Ka 3 = P[(Ka+Kb) / KaKb]

to find reaction F1= - Ka2= - Ka(P/Ka) = - P

finally the internal forces in the elements may be obtain using the

stiffness matrix for each element.

For element 1 :

=

aaa

aa

KpKK

kK

F

F

/

0.

2

1

F1 = - Ka * p/Ka = - p

F2 = Ka * p/Ka = pKa

P P


9/20

Example 2:The axially loaded bar system shown in Fig. below is formed from

three different segments. Applied loads of 100 and 150 kN are acted at

points 2 &3. Compute the displacements at these two nodes. Then

determine the reactions at the fixed ends of the system and draw the

axial force diagram. Use E= 30 kN/mm2

.

For element A :

=

2

1

2

1.

aa

aa

KK

kK

F

Fwhere : K=AE/L

=

2

1

2

1.

11

11

L

AE

F

F

mmkNL

EAK

a

aa

a/120

1000

30*10*40 2===

=

2

1

2

1.

120120

120120

F

F

For element B :

=

3

2

3

2.

bb

bb

KK

kK

F

F

=

3

2

3

2.

9090

9090

F

F

For element C :

=

4

3

4

3.

cc

cc

KK

kK

F

F

1

1000 mm

A=40*102

mm2Element 1

150 kN

2 3

4

A=60*102

mm2

2000 mm

A=40*102

mm2

100 kN

A B C

2000 mm

mmkNL

EAK

b

bbb

/902000

30*10*60 2===


10/20

=

4

3

4

3.

6060

6060

F

F

overall matrix

=

=

=

=

+

+

=

=

=

=

=

0

?

?

0

.

606000

606090900

09090120120

00120120

?

150

100

?

4

3

2

1

4

3

2

1

F

F

F

F

=

3

2.

15090

90210

150

100

The solution of this pair of simultaneous equations is :

mm

=

731.1

218.1

3

2

Now, we can find the reactions at nodes 1 &4 (F1 and F4 ).

=

=

=

=

=

0

731.1

218.1

0

.606000

00120120

4

3

2

1

4

1

F

F

kNF

F

=

86.103

16.146

4

1

The internal forces in the elements are:

For element A:

kNF

FA

=

=

16.146

16.146

218.1

0.

120120

120120

2

1

mmkNL

EAK

c

cc

c/60

2000

30*10*40 2===


11/20

146.1646.16

100

For element B:

kNF

FB

=

=

16.46

16.46

731.1

218.1.

9090

9090

3

2

For element C:

kNF

FC

=

=

84.103

84.103

0

731.1.

6060

6060

4

3

146.16 46.16 103.84103.84

150

Tension

compression

146.16

46.16

0

-103.84

Axial Force diagram


12/20

Example 3:Analyze the linear spring system shown in Fig. below ,determine the

displacements of nodes 2 and 3 , also find the reactions at nodes 1 and 4 .

where : K2 =K4 = K and K3 =2K K1= K5 = K/2 .

First we put the forces and displacements at the positive direction in

each node .

The element stiffness matrix for each element can be written as :


=

3

1

11

11

3

1.

KK

kK

F

F

=

2

1

22

22

2

1.

kK

kK

F

F

1

K1

K2F2

2

3 4K3F1K5

F3

F4

K4

1

K1

K2

2P

2

3 4K3

K4

K5

P


13/20


=

3

2

33

33

3

2.

KK

kK

F

F

=

4

2

44

44

4

2.

kK

kK

F

F

For element 5 :

=

4

3

55

55

4

3.

KK

kK

F

F

overall stiffness matrix for all system (assemblage of system).

+

++

++

+

=

4

3

2

1

5454

553131

434322

1221

4

3

2

1

.

0

0

KKKK

KKKKKK

KKKKKK

KKKK

F

F

F

F

For this system, nodes 1 and 4 are fixed (1 = 4=0 ). Also contain twounknown displacements (2 and 3 ) and unknown reaction forces (F1,F4).

F2 = -p

F3 = 2P

++

++=

3

2

5313

3432.

2 KKKK

KKKK

P

P

=

3

2.

32

24

2 KK

KK

P

P


14/20

-P = 4K2 2K32P = -2K2 + 3K3

solve the above equations to get the value of 2 and 3

and

The ratio of61

5.12*

83

2== P

KKP

Now To find the reactions (F1 and F4 )

8

7

2

5.1

81

P

K

P

KK

P

KF =

=

244 = KF

884

P

K

PKF =

=

K

P

82 =

K

P

2

5.13 =

31221 = KKF


15/20

Pin-Jointed Plane Bar Element:

The pin-jointed bar or truss element shown in Fig. below is the simplest type of

truss element. The bar element is assumed to have constant cross-sectional area (A)

and modulus of elasticity (E) over its length (L), also assumed to be homogeneous

and isotropic.

The stiffness matrix based on local coordinate system for this type

of elements as follows:

[ ]

=

0000

0101

0000

0101

L

AEK

e

For element :

{ } [ ] { }eee KF =

Global axis

Y

Local axis

Y'

F1'

F2'

F4'

F3'

X'

X

1

2

DOF= 4


16/20

In other terms, We rewrite the equation as:

=

4

3

2

1

4

3

2

1

0000

0101

0000

0101

L

AE

F

F

F

F

For transformation of the stiffness matrix based on global coordinate

system, referring to the above Fig. we will seen that :

sincos 433 FFF +=

cossin 434 FFF +=

similar equations hold at node 1. if we put = cos and = sin ,the above set of four equations may be written as :

=

4

3

2

1

4

3

2

1

00

00

00

00

F

F

F

F

F

F

F

F

OR

[ ]

=

00

00

00

00

T

{ } [ ]{ }FTF =


17/20

also :

{ } [ ]{ }= T

{ } [ ] { }= 1T

[ ] [ ] [ ] [ ]TKTK = 1

the stiffness matrix based on global coordinate system for bar

(truss) element :

[ ]

=

22

22

22

22

scsscs

csccsc

scsscs

csccsc

L

AEK

e

where C =cos and S = sin

Example 4:

Using the stiffness displacement method to find displacements at node

1 and hence solve for all the internal member forces and support

reactions.

Given:

E = 200 kN/mm2

A = 6400 mm2

45o

60o

40 kN

3 m

1

2 3 4

1

23

F1

F2

F3

F4

F5

F6

F7

F8


18/20

ELE. NODE O

COS SIN C.S C2

S2

L MM AE/L

1 1-2 135 -0.707 0.707 -0.5 0.5 0.5 4242 301.74

2 1-3 90 0 1 0 0 1 3000 426.67

3 1-4 30 0.866 0.5 0.433 0.75 0.25 6000 213.33

For Ele. 1

=

4

3

2

1

4

3

2

1

.

87.15087.15087.15087.150

87.15087.15087.15087.150

87.15087.15087.15087.150

87.15087.15087.15087.150

F

F

F

F

For Ele. 2

=

6

5

2

1

6

5

2

1

.

67.426067.4260

0000

67.426067.42600000

F

F

FF

For Ele.3

=

8

7

2

1

8

7

2

1

.

33.5337.9233.5337.92

37.9216037.92160

33.5337.9233.5337.9237.9216037.92160

F

F

FF


19/20

Overall stiffness matrix

=

8

7

6

5

4

3

2

1

8

7

6

5

4

3

2

1

33.5337.92000033.5337.92

37.92160000037.92160

0067.42600067.4260

00000000

000087.15087.15087.15087.150

000087.15087.15087.15087.15033.5337.9267.426087.15087.15087.63048.58

37.921600087.15087.15048.5887.310

F

F

F

F

F

FF

F

3 = 4 =5 = 6 =7 =8 = 0F1= 0 and F2 = -40

=

2

1

87.63048.58

48.5887.310

40

0

21 48.5887.3100 = 21 87.63048.5840 +=

solve (1) and (2) to get :

1= - 0.012142= - 0.06453

To find the reactions :

=

06453.0

01214.0

33.5337.92

37.92160

67.4260

00

87.15087.150

87.15087.150

8

7

6

5

4

3

F

F

F

F

F

F


20/20

F3 = -7.903 kN

F4 = 27.53 kN

F5 = 0

F6 = 27.53F7 = 7.903

To find the internal forces, first we obtain the local displacement due to

global displacement for each element.

For element one

{ } [ ] { }eee T =

=

=

=

=

=

0

0

0645.0

01214.0

.

707.0707.000

707.0707.000

00707.0707.0

00707.0707.0

4

3

2

1

4

3

2

1

0

0053.0

037.0

4

3

2

1

=

==

=

{ } [ ] { }eee KF =

=

=

=

=

=

0

0

053.0

037.0

.

0000

074.301074.301

0000

074.301074.301

4

3

2

1

4

3

2

1

F

F

F

F

Tension member

Similarly for elements 2 &3 .

0

164.11

0

164.11

3

2

1

=

=

=

=

F

F

F

F

F1/

1

2

X/

Y/

F3/

advanced structure

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