chapter 3: direct stiffness method

7/28/2019 Chapter 3: Direct Stiffness Method

1/30

University of California, San Diego Department of Structural EngineeringSE 201 Fall 2004 Class Notes Instructor: J oel P. Conte

1 1F ,

2 2F ,

3 3F ,

i

4 4F ,

5 5F ,

6 6F ,

j

1 1F ,

2 2F ,

3 3F ,

i

4 4F ,

5 5F ,

6 6F ,

j

2

1

3

NOT90

2

1

3

NOT90

3 DIRECT STIFFNESS METHOD APPLIED TO 2-D FRAME STRUCTURES

3.1 GENERAL

, :

F

= F k (3.1)

Figure 3.1

, : F

= F k (3.2)

Figure 3.2


2/30

3-2

4 4F ,

5 5F ,

6 6F ,

j

J

1 1F ,

2 2F ,

3 3F ,

i

I

4 4F ,

5 5F ,

6 6F ,

j

J

4 4F ,

5 5F ,

6 6F ,

j

JJJ

1 1F ,

2 2F ,

3 3F ,

i

I

1 1F ,

2 2F ,

3 3F ,

i

II

, :F = F k (3.3)

Figure 3.3

, :F

= F k (3.4)

Figure 3.4

4 4F ,

5 5F ,

6 6F ,

j

2xd

2yd

J

Y

X

E,G,A,

As,I,L

1 1F ,

2 2F ,

3 3F ,

i

1xd

1yd

I

4 4F ,

5 5F ,

6 6F ,

j

2xd

2yd

J 4 4F ,

5 5F ,

6 6F ,

j

2xd

2yd

JJ

Y

X

E,G,A,

As,I,L

1 1F ,

2 2F ,

3 3F ,

i

1xd

1yd

I

Y

X

Y

X

E,G,A,

As,I,L

E,G,A,

As,I,L

E,G,A,

As,I,L

1 1F ,

2 2F ,

3 3F ,

i

1xd

1yd

I1 1F ,

2 2F ,

3 3F ,

i

1xd

1yd

II


3/30

3-3

3.1.1 Structure Stiffness Matrix: K

Element a:

Node I = node 1

Node J = node 2

(a ) (a) (a )x x (node J) x (node I) 8 = - =

(a) (a ) (a )y y (node J) y (node I) 0 = - =

( ) ( )2 2

(a) (a ) (a )L x y 8= + =

(a )(a )

(a )

xcos 1

L

= =

(a )(a )

(a )

ysin 0

L

= =

Figure 3.5

(a)EI

= 25,000 k-inL

(a )EA k

= 3,750L in

Element b:

Node I = node 3

Node J = node 2

(b) (b) (b)x x (node J) x (node I) 8 = - =

(b) (b) (b)y y (node J) y (node I) 0 = - =

(b) 2 2L 8 6 10= + =

(b )8 4

cos 0.8010 5 = = =

(b )6

sin 0.6010

= =

(b)EI

165,000 k-inL

=

;

(b)EA k

4,200L in

=

Y

X

100 k

8

6

1 2a

b

3


4/30

3-4

Member a:

[ ]EI

25,000 k-inL

=

[ ]2EI 25,000 8 12 2,400,000 k-in= =

[ ]42,400,000

I 82.759 in29,000

= =

EA k3,750

L in

=

[ ]EA 3, 750 8 12 360, 000 k = =

[ ]2360,000

A 12.414 in29,000

= =

Member b:

[ ]EI

165,000 k-inL

=

[ ]2EI 165,000 10 12 19,800,000 k-in= =

[ ]419,800,000

I 682.759 in29,000

= =

EA k4,200

L in

=

[ ]EA 4, 200 10 12 504, 000 k = =

[ ]2504,000

A 17.379 in29,000

= =


5/30

3-5

1 2

3

1U

2U

3U

4U

5U

6U

7U

8U

9U

a

b

11 22

33

1U

2U

3U

4U

5U

6U

7U

8U

9U

a

b

3.1.2 Structure Degrees of Freedom Numbering:

Node 2 : free node

Node 1 and 3 : restrained nodes

Figure 3.6

Node displacements at free dofs: [ ]T

f 1 2 3U U U=U

Node displacements at restrained dofs: [ ]T

d 4 5 6 7 8 9U U U U U U=U


6/30

3-6

a

(a)

2F

(a)

1F

(a)

3F

(a)

5F

(a)

6F

(a)

4F1

6P

5P

7P

(a)

2F

(a)

1F

(a)

3F

21

P

2P

3P

(a)

5F(a)

6F

(a)

4F

(b)

5F

(b)

6F

(b)

4F

b

(b)

5F

(b)

6F

(b)

4F

(b)

2F

(b)

1F(b)

3F

37

P

9P

8P (b)2F

(b)

1F(b)

3F

a

(a)

2F

(a)

1F

(a)

3F

(a)

5F

(a)

6F

(a)

4F11

6P

5P

7P

(a)

2F

(a)

1F

(a)

3F

221

P

2P

3P

(a)

5F(a)

6F

(a)

4F

(b)

5F

(b)

6F

(b)

4F

b

(b)

5F

(b)

6F

(b)

4F

(b)

2F

(b)

1F(b)

3F

337

P

9P

8P (b)2F

(b)

1F(b)

3F

3.1.3 Equilibrium Equations for Free DOFs only:

[ ]T

f 1 2 3P P P=P : Applied forces

in free dofs( ) ( )

1 4 4P F Fa b= + ( ) ( )

2 5 5P F Fa b= +

( ) ( )

3 6 6P F Fa b

= +

[ ]T

d 4 5 6 7 8 9 10P P P P P P P=P :

forces at dofs with known

displacements (support reactions)

Figure 3.7

3.1.4 Direct Assembly/Formation of Equilibrium Matrix

Equilibrium equations for free and restrained dofs:

( ) ( )

1 4 4P F Fa b= + ( ) ( )

2 5 5P F Fa b= +

( ) ( )

3 6 6P F Fa b= +

( )

4 1P Fa=

( )

5 2P Fa= : Equilibrium between internal (Fs) and external (Ps) forces

( )

6 3P Fa=

( )

7 1P F b=

( )

8 2P Fb=

( )

9 3P Fb=


7/30

3-7

In matrix form:

(3.5)

Notes:

( )T( ) ( )f bf

ee e= P A F ( )T( ) ( )d bdee e= P A F

( ) ( )

f bf f

e e= A U ( ) ( )( ) f de ee += ( ) ( )d bd de e= A U

From element equilibrium, we can express the complete set of element end forces in the

global reference system 1 2 3 4 5 6( , , , , , )F F F F F F in terms of the basic forces 1 2 3( , , )

F F F as

T T TREZ ROT RBM

= F F

RBM ROT REZ( ) =

1

( )2

1

3

2

4f 3

5d 4

6

5

7

6

8

9

P 0 0 0 1 0 0 0 0 0 1 0 0P 0 0 0 0 1 0 0 0 0 0 1 0

FP 0 0 0 0 0 1 0 0 0 0 0

FP 1 0 0 0 0 0

FP 0 1 0 0 0 0

FP 0 0 1 0 0 0

FP 0 0 0 0 0 0

FP 0 0 0 0 0 0

P 0 0 0 0 0 0

a

= = = +

PP

P

( )

1

2

3

4

5

6

F1

F0 0 0 0 0 0

F0 0 0 0 0 0

F0 0 0 0 0 0

F1 0 0 0 0 0

F0 1 0 0 0 0

0 0 1 0 0 0

b

( )T

f

a= A ( )Tfb= A

( )T

d

a= A ( )Tdb= A


8/30

3-8

HereREZ = I (identity matrix, since = )

ROT

x y0 0 0 0

L L

y

x 0 0 0 0L L

0 0 1 0 0 0

x y0 0 0 0

L L

y x0 0 0 0

L L

0 0 0 0 0 1

=

RBM

1 10 1 0 0

L L

1 10 0 0 1

L L

1 0 0 1 0 0

=

2 2

2 2

T T T TREZ ROT RBM

2 2

2 2

y y x

L L L

x x y

L L L

1 0 0

y y x

L L L

x x y

L L L

0 1 0

= =


9/30

3-9

Thus,

( )f( )T ( )T

d

0 0 0 1 0 0 0 0 0 1 0 0

0 0 0 0 1 0 0 0 0 0 1 0

0 0 0 0 0 1 0 0 0 0 0 11 0 0 0 0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 0 0 1 0 0 0

aa b

= = +

PP F

P

( )bF


10/30

3-10

3.1.5 Direct Assembly/Formation of Compatibility Matrix

Compatibility between node displacements in global reference system and element end

displacements in global reference system:

Figure 3.8

(a) (b)

1 4 4U = =

2a

b

1

3

(a) (b)2 5 5U = =

2a

b

1

3

(a) (b)3 6 6U = = 2

a

b

3

1

(a) (b)3 6 6U = =

(a) (b)2 5 5U = =

(a) (b)1 4 4U = =


11/30

3-11

( )

1

2

1

3( ) ( )( )f 2 bf f

4

35

6

0 0 0

0 0 0U

0 0 0U

1 0 0U

0 1 0

0 0 1

a

a aa

= = = =

A U

N

( )

1

2

1

3( ) ( )( )f 2 bf f

4 Boolean displacement3 transformation matrix

5

6

0 0 0

0 0 0U

0 0 0U

1 0 0U

0 1 0

0 0 1

b

b bb

= = = =

A U

Using element compatibility, we can related the element deformations 1 2 3( , , ) to the

element end displacements in the global reference system 1 2 6( , , ... , ) :

( )( ) ( )

aa a = ; ( ) ( ) ( )b b b =

whereRBM ROT REZ REZ

(here )= = I

2 2 2 2

2 2 2 2

y x y x1 0

L L L L

y x y x0 1

L L L L

x y x y0 0

L L L L

=

( ) ( )( )f f

a aa = A U

( )( )( )

f f

bbb = A U


12/30

3-12

Form

( )

f f( )

a

b

= =

A U

where

( )( )

bff ( )( )

bf

:

aa

bb

=

AA

Astructure compatibility matrix for free dofs


13/30

3-13

3.1.6 Direct Stiffness Implementation

In the direct stiffness implementation of the displacement method, we proceed as follows:

We realize that the equilibrium equations at the free dofs can be written as (see p. 3-9):

( ) ( )( )T ( )T( )T ( )Tf bf bf

a ba ba b = + P A F A F

We express the basic element forces F in terms of basic element deformations according to the force-deformation relation of element e:

( ) ( ) ( )e e e = F k

Finally, the basic element deformations can be expressed in terms of the displacements at

the free global dofs according to (see p. 3-11):

( )( )( )

f f

eee = A U

In the absence of initial forces, we obtain

( ) ( ) ( ) ( )( )T ( )T( )T ( )T

f bf bf

a a b ba ba b = + P A k A k

( ) ( )( )T ( ) ( )T ( )( )T ( ) ( )T ( )

f bf bf f bf bf f a b

a a b ba a b b = + P A k A U A k A U

We note that the product

( )( )T ( ) ( )

aa a a = k k

is the element stiffness matrix in the global reference system.

After factoring out the displacements at the free global dofs, we obtain

( )T ( ) ( )T ( )( ) ( )f bf bf bf bf f

a a b ba b = + P A k A A k A U

We recognize in the square brackets the structure stiffness matrix for the free global dofs,

and generalizing toNelelements, we have

Nel( )T ( )( )

ff bf bf

1

e ee

e=

= K A k A (3.6)


14/30

3-14

3.1.7 Direct Assembly of Structure Stiffness Matrix

FromNel

( )T ( )( )ff bf bf

1

e ee

e=

= K A k A , we recognize that the structure stiffness matrix is derived

from the summation of element stiffness contributions, as long as these are expressed first

in the global reference system.

We also recognize that the compatibility matrices ( )fe

A are Boolean matrices of 1s and

0s. The 1s lie at the rows and columns corresponding to the relation between local

(element) and global (structure) dofs (each row and column thus contains at most one

non-zero term). Consequently, multiplication by ( )fe

A involves the positioning of the

terms of the element stiffness matrix in the appropriate address of the structure stiffness

matrix. Element contributions are summed up in the process. We illustrate this for

element a in the example:

which indicates that element dof 4 corresponds to global (structure) dof 1,

element dof 5 to global dof 2, and element dof 6 to global dof 3. The

other element dofs correspond to restrained global (structure) dofs.

ID array for element a reads

4 0

5 06 0

1 1

2 2

3 3

( )

bf

0 0 00 0 0

0 0 0

1 0 0

0 1 0

0 0 1

a

=

A


15/30

3-15

K( , ) K( , ) khid id id id = +

structure stiffness matrix element stiffness matrix

K( , ) K( , ) khid id id id = +

structure stiffness matrix element stiffness matrix

Similarly, ID array for element b reads

7 0

8 0

9 0

1 1

2 23 3

The ID array of the element provides the scheme for the proper addressing of the element

stiffness coefficients into the structure stiffness matrix as follows:

Note: In Matlab (Fedeaslab), the redirection of the stiffness coefficients of a single element

into the appropriate addresses of the structure stiffness matrix can be accomplished with

vector indexing in a single command:

11 12 13

ff 21 22 23

31 32 33

K K K

K K K

K K K

=

K

ID [ ]( ) ( ) ( ) ( ) ( ) ( )

11 12 13 14 15 16

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

21 22 23 24 25 26

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

31 32 33 34 35 36( )( ) ( ) ( ) ( ) ( )

41 42 43 44 45 46

0 0 0 1 2 3

0 k k k k k k

0 k k k k k k

0 k k k k k k

1 k k k k k k

2

3

a a a a a a

a a a a a a

a a a a a a

aa a a a a

=

k( )

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

51 52 53 54 55 56

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

61 62 63 64 65 66

k k k k k k

k k k k k k

a

a a a a a a

a a a a a a

11 12 13

ff 21 22 23

31 32 33

K K K

K K K

K K K

=

K

ID [ ]( ) ( ) ( ) ( ) ( ) ( )

11 12 13 14 15 16

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

21 22 23 24 25 26

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

31 32 33 34 35 36( )( ) ( ) ( ) ( ) ( )

41 42 43 44 45 46

0 0 0 1 2 3

0 k k k k k k

0 k k k k k k

0 k k k k k k

1 k k k k k k

2

3

a a a a a a

a a a a a a

a a a a a a

aa a a a a

=

k( )

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

51 52 53 54 55 56

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

61 62 63 64 65 66

k k k k k k

k k k k k k

a

a a a a a a

a a a a a a


16/30

3-16

3.1.8 Direct Assembly of Resisting Force Vector

Equilibrium equations (at the structure level):

r f=P P : express static equilibrium between external forces and internal (resisting) forces.

where r fResisting force vector in free global dof's=P U

f fApplied (external) forces in free global dof's=P U

( )T ( )T( ) ( )r bf bf

a ba b= + P A F A F

Generalizing toNelelements in the model:

Nel( )T ( )

r bf

1

e e

e=

= P A F (3.7)

It is important to recall that we have not yet included the initial end forces (fixed-end forces)

due to temperature changes, shrinkage, lack-of-fit (unintentional deviations from the reference

geometry), prestressing, etc. Well include these effects later.

( )

( ) ( )( ) T T T ( )T

REZ ROT RBM

e T

e ee e

=

= =

F F F

From element force-deformation relation:

( ) ( ) ( )e e e = F k where ( ) ( )( ) ( ) ( ) f fe ee e e = = A U

( )( )( ) ( )T ( )

f fe

ee e e = F k A U


17/30

3-17

3.1.9 Implementation of Direct Assembly

From the preceding discussion, it is apparent that it is possible to directly assemble the

structure resisting force vector (and thus express the structure equilibrium equations) and the

structure stiffness matrix. These two operations can be expressed in compact form as follows.

Nel( )T ( )( )

ff bf bf

1

e ee

e=

= K A k A : structure stiffness matrix (3.6)

where( )

( ) T T TREZ ROT RBM RBM ROT REZ

ee = k k

( )Te= ( )e=

( )( ) T T T

REZ ROT RBM RBM ROT REZ

ee = k k

( )Te= ( )e=

Nel( )T ( )

r bf

1

e e

e=

= P A F (3.7)

where, for linear elastic material response,

( )( )( ) ( )T ( )

f f

( )( )bf f

eee e e

ee

=

=

F k A U

k A U

r ff f = P K U : Structure resisting force vector

The structure equilibrium equations read: f r=P P

3.1.10 Structure State Determination

The process of determining the structure stiffness matrix ffK and the structure resisting force

vector rP is called structure state determination. During this process, each element e in the

model is called to supply its stiffness matrix ( )ek and resisting force vector ( )eF in global

coordinates. This sub-process is known as element state determination.


18/30

3-18

Back to the example:

Element a: units: k-in

( )

REZ

a = I (no rigid end-offsets)

( )

ROT

0 0 0 0 1 0 0 0 0 0

- 0 0 0 0 0 1 0 0 0 0

0 0 1 0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 1 0 0

0 0 0 - 0 0 0 0 0 1 0

0 0 0 0 0 1 0 0 0 0 0 1

cos 1

sin 0

a

c s

s c

c s

s c

c

s

= = =

= =

= =

I

RBM

( )

( )

1 11 10 1 0 00 1 0 0

96 96L L

1 1 1 10 0 0 1 0 0 0 1

L L 96 96

1 0 0 1 0 0 1 0 0 1 0 0

4 20

100,000(k-in) 50,000(k-in) 02 4

0 50,000(k-in) 100,000(k-in) 0

0

0 0

a

a

EI EI

L L

EI EI

L L

EA

L

= =

= =

k

0 3,750(k/in)

8 ft 96 inL = =

RBM

( )

( )

1 11 10 1 0 00 1 0 0

96 96L L

1 1 1 10 0 0 1 0 0 0 1

L L 96 96

1 0 0 1 0 0 1 0 0 1 0 0

4 20

100,000(k-in) 50,000(k-in) 02 4

0 50,000(k-in) 100,000(k-in) 0

0

0 0

a

a

EI EI

L L

EI EI

L L

EA

L

= =

= =

k

0 3,750(k/in)

8 ft 96 inL = =


19/30

3-19

( )( )T ( )T ( )T ( ) ( ) ( )( )REZ ROT RBM RBM ROT REZ

aa a a a a aa = k k

[ ]0 0 0 1 2 3

0 3,750. 0 0 3,750. 0 0

0 0 35.552 1,562.5 0 35.552 1,562.5

0 0 1,562.5 100,000. 0 1,562.5 50,000.

1 3,750. 0 0 3,750. 0 0

2 0 32

3

=

.552 1,562.5 0 32.552 1,562.5

0 1,562.5 50,000. 0 1,562.5 100,000.

11to K 22to K

23to K

ID array

of element a [ ]0 0 0 1 2 3

0 3,750. 0 0 3,750. 0 0

0 0 35.552 1,562.5 0 35.552 1,562.5

0 0 1,562.5 100,000. 0 1,562.5 50,000.

1 3,750. 0 0 3,750. 0 0

2 0 32

3

=

.552 1,562.5 0 32.552 1,562.5

0 1,562.5 50,000. 0 1,562.5 100,000.

11to K 22to K

23to K

ID array

of element a

ID array:

( )

1

( )

2

( )

3

( )

4 1

( )

5 2

( )

6 3

0

0

0

U

U

U

a

a

a

a

a

a

=

=

=

=

=

=

( )

1

( )

2

( )

3

( )

4 1

( )

5 2

( )

6 3

0

0

0

U

U

U

a

a

a

a

a

a

=

=

=

=

=

=

ID array of element a

0

0

0

1

2

3

=

Element b:

( )

REZ

b = I (no rigid end-offsets)

( )

ROT

0 0 0 0 0.8 0.6 0 0 0 0

- 0 0 0 0 0.6 0.8 0 0 0 0

0 0 1 0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 0.8 0.6 00 0 0 - 0 0 0 0 0.6 0.8 0

0 0 0 0 0 1 0 0 0 0 0 1

cos 0.80

sin 0.60

b

c s

s c

c ss c

c

s

= =

= =

= =


20/30

3-20

[ ]0 0 0 1 2 3

0 2,737.5 1,950. 4,950. 2,737.5 1,950. 4,950.

0 1,950. 1,600. 6,600. 1,950. 1,600. 6,600.

0 4,950. 6,600. 660,

1

2

3

=

000. 4,950. 6,600. 330,000.

2,737.5 1,950. 4,950. 2,737.5 1,950. 4,950.

1,950. 1,600. 6,600. 1,950. 1,600. 6,600.

4,950. 6,600. 330,000. 4,950. 6,600. 660,000.

11to K 22to K 23to K

ID array

of element b [ ]0 0 0 1 2 3

0 2,737.5 1,950. 4,950. 2,737.5 1,950. 4,950.

0 1,950. 1,600. 6,600. 1,950. 1,600. 6,600.

0 4,950. 6,600. 660,

1

2

3

=

000. 4,950. 6,600. 330,000.

2,737.5 1,950. 4,950. 2,737.5 1,950. 4,950.

1,950. 1,600. 6,600. 1,950. 1,600. 6,600.

4,950. 6,600. 330,000. 4,950. 6,600. 660,000.

11to K 22to K 23to K

ID array

of element b

( )bRBM

( )

( )

1 11 10 1 0 00 1 0 0

120 120L L

1 1 1 10 0 0 1 0 0 0 1

L L 120 120

1 0 0 1 0 0 1 0 0 1 0 0

4 20

660, 000(k-in) 330, 000(k-in) 02 4

0 330,000(k-in) 660,000(

0 0

b

b

EI EI

L L

EI EI

L L

EA

L

= =

= =

k k-in) 0

0 0 4,200(k/in)

10 ft 120 inL = =

( )bRBM

( )

( )

1 11 10 1 0 00 1 0 0

120 120L L

1 1 1 10 0 0 1 0 0 0 1

L L 120 120

1 0 0 1 0 0 1 0 0 1 0 0

4 20

660, 000(k-in) 330, 000(k-in) 02 4

0 330,000(k-in) 660,000(

0 0

b

b

EI EI

L L

EI EI

L L

EA

L

= =

= =

k k-in) 0

0 0 4,200(k/in)

10 ft 120 inL = =

( )( )T ( )T ( )T ( ) ( ) ( )( )REZ ROT RBM RBM ROT REZ

bb b b b b bb = k k


21/30

3-21

Direct assembly of structure stiffness matrix:

Nel=2( )T ( )( )

ff bf bf

1

6,487.5 1,950. 4,950.

1,950. 1,632.552 8,162.5

4,950. 8,162.5 760, 000.

e ee

e=

=

=

K A k A

Equilibrium equations: f f f f =K U P

1

2

3

6, 487.5 1,950. 4,950. U 0

1,950. 1,632.552 8,162.5 U 100

4,950. 8,162.5 760,000. U 0

=

1

2

3

1

2

3

6, 487.5 1,950. 4,950. U 0

1,950. 1,632.552 8,162.5 U 100

4,950. 8,162.5 760,000. U 0

=

11

22

33

Eq. 11 expresses the horizontal equilibrium of node 2 (free node 1).

Eq. 22 expresses the vertical equilibrium of node 2 (free node 1).

Eq. 33 expresses the rotational equilibrium of node 2 (free node 1).

Solve equilibrium equations for fU (using Matlab)

-1

f ff f = U K P

0.03361 in

0.10831 in

0.001382 rad

=

Check thatf ff f

f

-1

P K U

P


22/30

3-22

3.1.10.1 Element State Determination (more generally known as stress recovery phase in finiteelement analysis)

Element a:

( )

1

2

3

0 0

0 0

0 0

U 0.03361

U 0.10831

U 0.001382

a

= =

( ) ( ) ( ) ( ) ( )RBM ROT REZ

0.0011282 rad

0.00025394 rad

0.033610 in

a a a a a

= =

( ) ( ) ( ) ( )

100,000. 50,000. 0

50,000. 100,000. 0

0 0 3,750.

a a a a

= =

F k

( )

I

( )

J

( )

100.12 M (k-in)

31.02 M (k-in)

126. N (k)

a

a

a

=

Element b:

( )

1

2

3

0 0

0 00 0

U 0.03361

U 0.10831

U 0.001382

b

= =


23/30

3-23

( )( ) ( ) ( ) ( )RBM ROT REZ

0.00089 rad

0.000492 rad

0.038098 in

bb b b b

= =

( ) ( ) ( ) ( )

660,000. 330,000. 0

330,000. 660,000. 0

0 0 4,200.

b b b b

= =

F k

( )

I

( )

J

( )

425.1 M (k-in)

31.02 M (k-in)

160. N (k)

b

b

b

=


24/30

3-24

3.1.10.2 Check Nodal Equilibr ium and Determine Support Reactions

Figure 3.9

x

4F 126 (160)

5

3(3.28) 0.032

5

0 (round-off error)

= +

= +

x4

F 126 (160)5

3(3.28) 0.032

5

0 (round-off error)

= +

= +

x

4F 100 1.366 (3.28

53

(160) 0.015

0 (round-off error)

= + +

+ =

x4

F 100 1.366 (3.28

53

(160) 0.015

0 (round-off error)

= + +

+ =

M 31.02 31.02

0

= +

=

2M 31.02 31.020

= +

=

2M 31.02 31.020

= +

=

22

Equilibrium Check

3

425

3.28

3.28

31.02

31.02

3.28

( )425 31.02

3.28 =10 12

16

0

160

160

10ft

160

3

4

5

( ) ( )4 3

160 3.285 5

126.03=

425

( ) ( )3 4

160 3.28 98.6245 5

+ =

Support reactions

at Node 3

1001.366 31.02

1262

126

1.366

100.12 1.366

126

100.12

1

( )100.12+31.02

1.366 =8 12

126

100.12

126

31.02

1.366

8 ft

Support reactions

at Node 1

425

Units: k, in

Equilibrium Check

33

425

3.28

3.28

31.02

31.02

3.28

( )425 31.02

3.28 =10 12

16

0

160

160

10ft

160

3

4

5

( ) ( )4 3

160 3.285 5

126.03=

425

( ) ( )3 4

160 3.28 98.6245 5

+ =

Support reactions

at Node 3

1001.366 31.02

12622

126

1.366

100.12 1.366

126

100.12

11

( )100.12+31.02

1.366 =8 12

126

100.12

126

31.02

1.366

8 ft

Support reactions

at Node 1

425

Units: k, in


25/30

3-25

a

b

126

1.366

100.12

126.03

98.624

425

100

2

3

1a

b

126

1.366

100.12

126.03

98.624

425

100

22

33

11xF 126 126.03 0.03 0 (round-off error)= + = + xF 126 126.03 0.03 0 (round-off error)= + = +

yF 1.366 100 98.624 0.01 0 (round-off er= + = yF 1.366 100 98.624 0.01 0 (round-off er= + =

M 100.12 126(6)(12) 425 100(8)(12)

2.88 0 (round-off error)

= + + =

=

2M 100.12 126(6)(12) 425 100(8)(12)2.88 0 (round-off error)

= + + =

=

22

425

100.12

31.02

31.0

2

425

100.12

31.02

31.0

2100.12

31.02

31.0

2

3.1.10.3 Check Global Equilibrium

Figure 3.10

Draw internal forces (M,V, N) diagrams according to sign convention:

Bending moment diagram M:

Figure 3.11


26/30

3-26

1.371.373.28

3.28

1.371.373.28

3.28

2

3

1

2

inflection point

100 k

chord

+ 0.0011282 rad0.00025394 rad

0.000492 rad

0.00089 rad+

22

33

11

2

inflection point

100 k

chord

+ 0.0011282 rad0.00025394 rad

0.000492 rad

0.00089 rad+

Shear force diagram V:

Figure 3.12

Axial force diagram N:

16

0

12612616

0

16

0

12612616

0126126

16

0

Figure 3.13

Draw accurate sketch of deflected shape consistent with internal forces, especially bending

moment diagram:

Figure 3.14


27/30

3-27

166.67

166.67 166.67

( )5

1003

=

133.33

(5.8% difference with frame results)

133.334

= (100)

3

100

2

(4% difference with frame results)

3

166.67

133.33

100

133.331

0

166.67

166.67 166.67

( )5

1003

=

133.33

(5.8% difference with frame results)

133.334

= (100)

3

100

2

100

22

(4% difference with frame results)

33

166.67

133.33

100

133.3311

0

If the above structure were to resist as a truss:

Figure 3.15

Frame action is very modest in this structural system.


28/30

3-28

3.2 SLAVING

Slaving is a particular case of elements with Rigid End Zones. In this case, the dofs of

several nodes are slaved to those of a Master node. In other words, the displacements of

several nodes are rigidly linked to those of a single Master node. The typical example is that

of a rigid floor, where the displacements of the story columns are rigidly linked to those of therigid floor, represented by a Master node. In the figure below, the displacements at the top of

column i,( ) [ ]

Ti

i i iu v = , depend on (are rigidly linked to) those of Master node M

(typically taken as the center of stiffness or the center of mass of the floor),

[ ]T

M M M Mu v = .

uM

vM MMaster

node

Rigid floor (in- and

out-of-plane)

Axially

rigid

columns

Single story shear building model

uM

vM MMaster

node


out-of-plane)

Axially

rigid

columns

uM

vM MMaster

node


out-of-plane)

Axially

rigid

columns

Single story shear building model

vMM

x

y

uM

ui

vii

i yi

xi

vMM

x

y

uM

ui

vii

i yi

xi

For node i, whose coordinates are (xi, yi):

( ) ( )

1 0

0 1

0 0 1

i

i i

REZ M i M

y

x

= =

Based on the properties of the transformation matrices:

( ) ( ) ( )T

i i i

M REZ= F F


29/30

3-29

where( )i

F are the forces from the slab onto the top of column i and( )i

F represent the forces at

Master nodeM, which are in equilibrium with( )i

F . Thus:

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

( )

( )T T T

iM

i i i i i i i i i i

REZ REZ REZ REZ M M M= = = =

k

F F k k k

where( ) ( ) ( ) ( )

Ti i i i

REZ REZ= k k

After assembly of all the column contributions to the equilibrium equations of the slab, we

have:

M M =K F

where MF denote the external forces applied to Master nodeM, K is the stiffness of master

nodeM(or of the one-story building structure shown in the figure above), and

are thedisplacements of master nodeM.

General Procedure (for one-story shear building model):

(1) For each column element, find the element stiffness matrix in coordinates( )i

and ,

respectively:

( )ik ,

( ) ( ) ( ) ( )T

i i i i

REZ REZ= k k

(2) Assemble the Master node stiffness matrix:

( )_ _

1

number of columnsi

M M

i

k k=

=

(3) Find by solving:

M M= F K

(4) Find the top column displacements:

( ) ( )i iREZ M=

(5) Find the top column forces:

( ) ( ) ( )i i i= F k


30/30

3-30

General Case:

The general case is that of a multi-story shear

building model (rigid floors and inextensible

columns), as shown below. Each floor has 3dofs.

1

2 3

4

5 6

7

8 9

10

11 12

chapter 3: direct stiffness method

Documents