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Introduction to Computational Engineering - Supplement1

Example 1:

The structure shown consists of linear springs whose stiffnesssare 1 2 3 4, , andk k k k . Only horizontal displacements are allowed.

(a)In matrix, write the three equilibrium equations of thestructures. The d.o.f. are 1 2 3, andu u u .

(b)Let 1 2 3 4 1 2and 0.k k k k F F = = = = = Determine 1 2 3, andu u u interms of 3andk F .

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Solution:

1 1u , F

2 2u , F 3 3u , F

2k

3k

4k

1k

a.

( ) ( ) ( )

1

1 1 1 1 2 1 4 1

2 1 2 2

3 1 4 3

1 1 2 4 1 2 2 4 3

* Calculating

activating

activating

activatingBy superposition:

= + +

=

=

= + + + +

F

u F k u k u k u

u F k u

u F k u

F k k k u k u k u

( ) ( ) ( )

2

1 2 2 1

2 2 2 2 3 2

3 2 3 3

2 2 1 2 3 2 3 3

* Calculating

activating

activating

activating

By superposition:

F

u F k u

u F k u k u

u F k u

F k u k k u k u

=

= +

=

= + + +

( )

( ) ( ) ( )

3

1 3 4 1

2 3 3 2

3 3 3 4 3

3 4 1 3 2 3 4 3

* Calculating

activating

activating

activating

By superposition:

F

u F k u

u F k u

u F k k u

F k u k u k k u

=

=

= +

= + + +

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( )

( )( )

1 2 4 2 4 1 1

2 2 3 3 2 2

4 3 3 4 3 3

In matrix form:

k k k k k u F

k k k k u F

k k k k u F

+ + + = +

b.1

2

3 3

1 3

31

3 32 3

3 1 1 0

1 2 1 0

1 1 2

adding the three equations

substitute

4 5and3 3

u

k u

u F

ku F

Fu

k

F Fu uk k

=

=

=

= =

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Example 2:

Activate DOF in turn.

1 1k u

2 1k u

1M

1M

1M

1M

2M 2M

2M 2M

2F

2F

2F

2F

1F

1F

1F

1F

1 2k u

2 2k u

1 1k a

2 1k a

1 2k a

2 2k a

( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )

1 2 2 1 1 2 1 2

2 2

2 1 1 2 1 2 1 2

1 2 1 2 1 2 2 1

2 2

1 2 1 2 2 1 1 2

k k a k k k k a k k

a k k a k k a k k a k k

k k a k k k k a k k

a k k a k k a k k a k k

+ +

+ + + +

+ +

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Example 3:

The plane truss shown has bars of stiffness k1, k2, k3 and k4,

where ki =AiEi/Li.

I. Free all d.o.f and write the structure stiffness matrix interms of the ki. Let nodal d.o.f. have the order:

{ } 1 1 2 2 3 3 4 4T

D u v u v u v u v=

II. Considering the plane truss:a. Impose support conditions implied by the sketch.

That is, by discarding the appropriate rows andcolumns, obtain a smaller [K] that operates only theactive d.o.f.

b. For the loading by force F shown, write the 4 1vecore {D} by inspection (not by solving simultaneousequations). Hence, find the nodal loads {R} = [K]{D}.

Are these loads physically reasonable?

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Solution:

The system includes:

4 elements 4 nodes.

Each node has two degrees of freedom (d.o.f.s): ux, uyTotal number of d.o.f.s = 4 2 = 8

Active d.o.f.s: u2, u3, u4, v4

Restricted d.o.f.s: u1, v1, v2, v3.

[ ] [ ]

[ ] [ ]

1 1 2 2 3 3 4 4

1 2 3 4 5 6 7 8

'the displacement vector'

'the force vector'

D u v u v u v u v

Q Q Q Q Q Q Q Q Q

=

=

Consider node

Free body diagram (FBD)

Free Body Diagram

We consider the nodal loads consistent with the followingdisplacement states:

[ ] [ ]

[ ] [ ]

1

1

0 0 0 0 0 0 0

0 0 0 0 0 0 0

D u

D v

=

=

Let [Q1] represent the vector of forces associated with unitdisplacement u1 = 1:

[ ] [ ]1 1 4 4 10 0 0 0 0 0T

Q u k k u=

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Similarly [Q2] is the force vector associated with the unit

displacement v1 = 1:

[ ] [ ]2 1 10 0 0 0 0 0 0 0T

Q v v=

Approach:

Let [ ] [ ] [ ] [ ] [ ] [ ]3 4 5 6 7 8, , , , , andQ Q Q Q Q Q represent theequilibrium nodal force vectors associated with the remainingsix unit displacement states (u2 = 1, v2 = 1, u3 .). Then, if all

eight nodal d.o.f.s may be non-zero simultaneously, bysuperposition:

[ ]

1 1

1 1

2 2

2 2

1 2 3 4 5 6 7 8

3 3

3 3

4 4

4 4

u p

v q

u p

v qQ Q Q Q Q Q Q Q

u p

v q

u p

v q

=

Consider node

We consider the nodal loads consistent with the respective statesof displacement.

[ ] [ ]

[ ] [ ]

2

2

0 0 0 0 0 0 0 and

0 0 0 0 0 0 0

T

T

D u

D v

=

=

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Free Body Diagramdue to u2:

Elongation in

member = e1

e1 = u2 cos = 0.6 u2

2cosu

2u

2u

1e

[ ] ( )

1 1 1 1 2

1 1 1 2

1 1 1 2

2 2 2

3 3 3

4 4 4 2 4

3 4 4 1 1 1 1 2

0 6

0 6 0 36

0 8 0 48

0 0 0

0 0 0

0

0 0 36 0 48 0 36 0 48 0 0

x

y

x y

x y

x y

T

F e k . k u

F . F . k u

F . F . k u

F F F

F F F

F F k u F

Q k k . k . k . k . k u

= =

= =

= =

= = =

= = =

= = =

= +

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Free Body Diagramdue to v2:

Elongation in

member = e1

e1 = v2 sin = 0.8 v2

2u

1e

2

1 1 1 1 2

1 1 1 2

1 1 1 2

2 2 2 2

3 3

4 4

0 80 6 0 48

0 8 0 64

0

0 0

0 0

x

y

x y

x y

x y

F e k . k vF . F . k v

F . F . k v

F F k v

F F

F F

= == =

= =

= =

= =

= =

[ ] ( )4 1 1 2 1 1 20 0 0 48 0 64 0 48 0 64 0 0T

Q . k . k k . k . k v = +

Consider node

We consider the nodal loads consistent with the respectivedisplacement states:

[ ] [ ]

[ ] [ ]

T

3

T

3

0 0 0 0 0 0 0 and

0 0 0 0 0 0 0

D u

D v

=

=

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Free Body Diagramdue to u3:

Elongation in member

e1 = u2 cos = 0.6 u2

(1)

(3)3u

1 1 1 1 2

1 1 1 2

1 1 1 2

2 2

3 3 3 3

4 4

0 6

0 6 0 36

0 8 0 48

0 0

00 0

x

y

x y

x y

x y

F e k . k u

F . F . k u

F . F . k u

F F

F k u F F F

= =

= =

= =

= =

= == =

3u

1e

[ ] ( )4 1 1 1 3 1 3 30 0 0 36 0 48 0 36 0 48 0T

Q . k . k . k k . k k u = +

Free Body Diagramdue to v3:

Elongation in member= e1

e1 = v3 sin = 0.8 v3

(1)

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Free Body Diagramdue to u4:

1 1 1

2 2 2

3 3 4 3

4 4

0

0

0

0 0

x y

x y

x y

x y

F F F

F F F

F k u F

F F

= = =

= = =

= =

= =

34

21

4u

(1)

(4)

(3)

(2)

[ ] [ ]5 3 3 30 0 0 0 0 0T

Q k k v=

Free Body Diagramdue to v

4:

1 1 1

2 2 2 4

3 3 3

4 4 4

0

0

0

0

x y

x y

x y

x y

F F F

F F k v

F F F

F F F

= = =

= =

= = =

= = =

(1)

(3)

(4)

(2)

34

21

4

[ ] [ ]6 2 2 40 0 0 0 0 0T

Q k k v=

Assembly of the stiffness matrix: [K]

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4 4

4 4 1 1 1 1

1 2 1 1 1 2

1 1 1 3 1 3

1 1 1 1

3 3

2 2

0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 36 0 48 0 36 0 48 0 0

0 0 0 48 0 64 0 48 0 64 0

0 0 3 6 0 48 0 36 0 48 00 0 0 48 0 64 0 48 0 64 0 0

0 0 0 0 0 0

0 0 0 0 0 0

k k

k k . k . k . k . k

. k k . k . k . k k

. k . k . k k . k k

. k . k . k . k

k k

k k

+

+

+

The structure stiffness equations are:

[K][D] = [R]

where:[D] is the vector of nodal d.o.f.

[R] is the vector of nodal loads

[K] is the total stiffness matrix (as shown above)

The structural equations [K][D] = [R] can be written in the form:

11 12

21 22

x c

c x

D RK K

D RK K

=

where [Dc] and [R

c] are known and [D

x] and [R

x] are still

unknown.Expanding:

[ ][ ] [ ] [ ]

[ ][ ] [ ] [ ]

11 12

21 22

x c c

x c x

K D K D R

K D K D R

+ =

+ =

The first equation

[ ] [ ] [ ] [ ][ ]{ }1

11 12 x c c D K R K D

=

If the known DOFs (boundary conditions) have zero values:

[ ] [ ][ ] [ ] [ ]

1

11

0c

x c

D

D K R

==

where K11is the remainder of the matrix after discarding rows

and columns related to the zero d.o.f.s.

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In this example:

1 1 2 3 0u v v v= = = =

The discarded rows and columns will be:

4 4

4 4 1 1 1 1

1 2 1 1 1 2

1 1 1 3 1 3

1 1 1 1

3 3

2 2

0 0 0 0 0 00 0 0 0 0 0 0 0

0 0 36 0 48 0 36 0 48 0 0

0 0 0 48 0 64 0 48 0 64 0

0 0 3 6 0 48 0 36 0 48 0

0 0 0 48 0 64 0 48 0 64 0 0

0 0 0 0 0 0

0 0 0 0 0 0

k k

k k . k . k . k . k

. k k . k . k . k k

. k . k . k k . k k

. k . k . k . k

k k

k k

+

+ +

The stiffness matrix [K11] becomes:

[ ]

4 1 1

1 1 3 3

11

3 3

2

0 36 0 36 0 0

0 36 0 36 0

0 0

0 0 0

k . k . k

. k . k k k K

k k

k

+ + =

The structural equations become:

4 1 1 2

1 1 3 3 3

3 3 4

2 4

0 36 0 36 0 0

0 36 0 36 0 0

0 0 0

0 0 0 0

k . k . k u F

. k . k k k u

k k u

k v

+ + =

By inspection:

u2 = u3 = u4 = F/k4, v4 = 0

Bars 1, 2, 3 translate right the same amount as a rigid body.

[ ] [ ]4

1 1 1 0TF

Dk

=

For which, substituting in [R] = k[D]

[ ] [ ]0 0 0 0 0 0 R F F =

Bar 4 stretches the amount FL/AE = F/k4.

There are no reactions at supports 2 and 3.

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Example 4:

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Solution:

Change of angle at B is4

C Au u

L

Activate uAone unit, then uCone unit.

4

Bk

L4

Bk

L

CF

AF

CF

Ak

Ak

4

Bk

L4

Bk

L

CF

AF

CF

2 2

2

2 2

2 2

4 0;

4 16 16

16

16 16

16 16

B B BC C C

B A A C A A C

B BA

B B

k k k LF F F

L L Lk

F k F k F F L

k kk

L Lk

k k

L L

+ = = =

= = + =

+

=